Physics 1120: Centre of Mass, Torque, & Static
Equilibrium Solutions
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Centre of Mass
- The distance between the oxygen molecule and
each of the hydrogen
atoms in a water (H2O) molecule is 0.0958 nm; the angle
between the two oxygen-hydrogen bonds is 105°.
Treating the atoms as particles, find the centre of mass.
The problem expects you to recall that the mass
of an oxygen atom is 16 times that of a hydrogen atom. The first
step is to choose a coordinate system, such as the one in the
diagram, and locate each particle. The chosen origin is the centre
of the box.Atom Mass (H) xi yi mixi miyi H 1 -0.0958sin15 0.0958cos15 -0.02479 0.09254 O 16 0 0 0 0 H 1 0.0958 0 0.0958 0 Totals: 18 0.07101 0.09254 The coordinates of the centre of mass are given by
xcm = (Σmixi)/Mtotal
= 0.07101/18 = 0.0039 nm, andycm = (miyi)/Mtotal
= 0.09254/18 = 0.0051 nm.The centre of mass is located at (0.0039 nm, 0.0051 nm). Answers
will vary based on the choice of coordinate system.
- Where is the centre of mass of a uniform cubic box of side
length L which has no lid?In dealing with real objects rather than particles, we treat
the complex object as a grouping of simpler shapes. The CM of
the simpler shapes is at their easy to find geometric centre if
the object is uniform. Each pierce can now be considered a particle
with the mass of the piece located at the CM of that piece. We
have, in effect, turned the complex shape into a collection of
particles. In this case, each of the five sides can be considered
a separate particle.The next step is to choose a coordinate system, such as the one
in the diagram below, and locate each particle. The origin is
at the centre of the box.Side Mass xi yi zi mixi miyi mizi bottom M 0 0 -½L 0 0 -½ML front M 0 -½L 0 0 -½ML 0 back M 0 ½L 0 0 ½ML 0 left M -½L 0 0 -½ML 0 0 right M ½L 0 0 ½ML 0 0 Totals: 5M 0 0 -½ML The coordinates of the centre of mass are given by
xcm = (Σmixi)/Mtotal
= 0,ycm = (Σmiyi)/Mtotal
= 0, andzcm = (Σmizi)/Mtotal
= -½ML / 5M = -L/10.The centre of mass is located at (0, 0, -L/10). Answers will
vary based on the choice of coordinate system.It is also permissible to use symmetry arguments. For example,
the figure in the diagram is only unbalanced in the z direction,
thus we know that xcm = ycm = 0. We only
needed the z columns in the above table. - Two uniform squares of sheet metal of dimension L ×
L are joined at a right angle along one edge. One of the squares
has twice the mass of the other. Find the centre of mass.In dealing with real objects rather than particles,
we treat the complex object as a grouping of simpler shapes.
The CM of the simpler shapes is at their easy to find geometric
centre if the object is uniform. Each pierce can now be considered
a particle with the mass of the piece located at the CM of that
piece. We have, in effect, turned the complex shape into a collection
of particles. In this case, we have one particle of mass M located
in the centre of the lighter side, and a mass of 2M in the centre
of the heavier side.The next step is to choose a coordinate system,
such as the one in the diagram below, and locate each particle. The
origin is in the centre of the join of the two plates.Using the symmetry of the problem, we see that the
CM must be located in the xz plane, we know that
ycm = 0.Side Mass xi zi mixi mizi side M 0 ½L 0 ½ML bottom 2M ½L 0 ML 0 Totals: 3M ML ½ML The components of the centre of mass are given by
xcm = (Σmixi)/Mtotal
= ML / 3M = L/3,zcm = (Σmizi)/Mtotal
= ½ML / 3M = L/6.The centre of mass is located at (L/3, 0, L/6). Answers will
vary based on the choice of coordinate system.
- A cube of iron has dimension L ×
L × L. A hole of radius ¼L
has been drilled all the way through the cube, so that one side
of the hole is tangent to one face along its entire length. Where
is the centre of mass of the drilled cube?In dealing with real objects rather than particles,
we treat the complex object as a grouping of simpler shapes.
The CM of the simpler shapes is at their easy to find geometric
centre if the object is uniform. Each pierce can now be considered
a particle with the mass of the piece located at the CM of that
piece. We have, in effect, turned the complex shape into a collection
of particles. In this case, we have a solid cube and a cylindrical
hole. We treat holes as objects of negative mass.To proceed we need to know the mass of the cylindrical
hole. Since the object was uniform, its mass is proportional
to its volume. The solid cube had a mass M and a volume L3.
The cylinder has a volume Vcyl = πr2L =
πL3/16. Thus the mass of the cylindrical hole ismcyl = mcube[Vcyl
/ Vcube] = M[(πL3/16) / L3]
= πM/16.The next step is to choose a coordinate system,
such as the one in the diagram below, and locate each particle.Using the symmetry of the problem, we see that the
CM must be located in the x axis, we know that ycm
= zcm = 0.Side Mass xi mixi solid cube M 0 0 hole -πM/16 -¼L πML/64 Totals: M(1-π/16) πML/64 The location of the x component of the centre of mass is given
byxcm = (Σmixi)/Mtotal
= -πML/64 / -M(1-π/16)
= πL / 64(1-π/16).The centre of mass is located at (πL / 64(1-π/16), 0, 0).
Answers will vary based on the choice of coordinate system. - An 80-kg logger is standing on one end of a 10-m long, 300-kg,
tree trunk in the middle of the Fraser River. The logger walks
upriver along the trunk to the other end of the log. As a result
the log moves some distance L down river. What is the displacement
L?The logger and the log are a system; the system has a certain
centre of mass. The motion of the logger is an internal force;
internal forces cannot change the centre of mass of the system.Examining the diagram, the log has moved down the river a distance
equal to twice the distance between the centre of the log and
the CM of the logger-log system. We need to find the CM of the
logger-log system. Taking the origin as the end of the log,xcm = (Σmixi)/Mtotal
= [80×0 + 300×5] / 380 = 3.047 m.Since the centre of the log is at 5 m, distance between the CM
and the centre of the log is 1.05 m. The log moved twice this
distance or 2.11 m. - A shell is fired at 25 m/s at 25 above the horizontal. At
the top of its parabolic flight, it breaks into two pieces. One
piece, having two-thirds of the total mass of the shell lands
60 m from where the shell was fired. Where did the other piece
land?The pieces of the shell are a system; the system has a certain
centre of mass. The explosion is an internal force; internal
forces cannot change the centre of mass of the system. The CM
of the pieces will land where the CM of an unexploded shell will
land.The first step is to find xcm, the landing position
of the shell. That involves solving the projectile motion problem.i j x = xcm = ? y = 0 ax = 0 ay = -g = -9.81 m/s v0x = 25cos25 = 22.658 m/s v0y = 25sin25 = 10.565 m/s t = ? t = ? The j information allows us to find the time in
air using y = v0yt – ½gt2. Since y
= 0, this reduces tot = 2voy/g = 2×10.565 / 9.81 = 2.154 s.
We then find the landing position using x = v0xt +
½axt2. Since ax = 0,xcm = voxt = 22.658 × 2.154 =
48.806 m.The centre of mass is determined by the formula
xcm = (Σmixi)/Mtotal
= m1x1/Mtotal + m2x2/Mtotal.Since we now know xcm and x2, we can rearrange
to find m1,x1 = [Mtotalxcm – m2x2]/m1
= [1×48.806 – (2/3)60] / (1/3) = 26.4 m.So the smaller piece lands 26.4 m from where the shell was fired.
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Torque and Static Equilibrium
- In the diagram below, three forces are applied
to a 3-4-5 triangle. The forces are F1 = 91.7 N, F2
= 150 N, and F3 = 67.7 N. F3 is applied
at the middle of side AB. (a) Find the net torque about point
A. (b) Find the net torque about point B. (c) Find the net torque
about point C.There are two methods for determining torque. Method A is to
use τz
= rFsinθ, where r is the distance from the pivot
point to the point where the force F acts and
θ is the angle between
r and F. The sign of τz is found
using
the right-hand rule. Method B is to use τz = xFy
– yFx, where (x, y) is the location of where the force
is acting taken relative to the pivot point which is taken to
be the origin (0, 0). Fx and Fy are the
components of the force – careful attention must be paid to signs.Method A.
First note that the interior angles of the triangle are α
= tan-1(4/3) = 53.130°, and γ
= tan-1 (3/4) = 36.870°. F2 makes an angle
φ = 180° – 110°
– γ = 16.870° with the vertical.(a)
r (m) F (N) θ direction τz = rFsinθ (N-m) 0 91.7 – – 0 5 150 110° CCW 704.769 2 67.7 130° CW -103.722 Total: 601.0 (b)
r (m) F (N) θ direction τz = rFsinθ (N-m) 4 91.7 90° CCW 366.800 3 150 110° + γ CCW 130.591 2 67.7 50° CCW 103.722 Total: 601.1 (c)
r (m) F (N) direction τz = rFsinθ (N-m) 5 91.7 90° + α CCW 366.800 0 150 – – 0 3.60555 67.7 50° + 56.130° CCW 234.272 Total: 601.1 Method B:
(a)
x (m) y (m) Fx (N) Fy (N) τz = xFy – yFx
(N-m)0 0 0 -91.7 0 4 3 -150sinφ 150cosφ 704.769 2 0 67.7cos50° -67.7sin50° -103.722 total: 601.0 (b)
x (m) y (m) Fx (N) Fy (N) τz = xFy – yFx
(N-m)-4 0 0 -91.7 103.722 0 3 -150sinφ 150cosφ 130.591 -2 0 67.7cos50° -67.7sin50° 366.80 total: 601.1 (c)
x (m) y (m) Fx (N) Fy (N) τz = xFy – yFx
(N-m)-4 -3 0 -91.7 366.800 0 0 -150sinφ 150cosφ 0 -2 -3 67.7cos50° -67.7sin50° 234.272 total: 601.1 Please note that the only reason the total torque at point A,
B, and C are the same is because F1 + F2
+ F3 = 0.
- An L-shaped object of uniform density is hung
over a nail so that it is free to pivot. What angle, θ,
does the long side make with the vertical? The long side of the
L-shaped object is twice as long as the short side?The problem mentioned that the object is free to
pivot, to rotate. This indicates that we are dealing with a Static
Equilibrium problem. We solve Static Equilibrium problems by
sketching the extended free-body diagram, an FBD where the location
of the all forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the L-shaped
object are a normal from the nail and the weight which acts from
the centre of mass. Ordinarily, for complex shapes, we first
determine the CM. However, in this case, it is easier to consider
the two arms of the objects as being separate objects. The long
arm will have a mass (2/3)mg and the short arm will be (1/3)mg.
We do not have a simple method of figuring out which way the
normal points. As with all pins, we consider it as two forces
one vertical and one horizontal.ΣFx = 0 ΣFy = 0 Nx = 0 Ny – (1/3)mg – (2/3)mg = 0 These tell us the obvious, the normal has no horizontal component
and that it supports the weight of the object.We will use Method A for the torques since that method is
easiest
to apply here. We will take the nail as the pivot point since
this eliminates the torques from the nail.r F direction τz = rFsinθ 0 Nx – – 0 0 Ny – – 0 L/2 (1/3)mg ½π-θ CW -mgLsin(½π-θ)/6 L (2/3)mg θ CCW 2mgLsinθ/3 Since Στz = 0, the equation we get is
-mgLsin(½π-θ)/6
+ 2mgLsinθ/3 = 0.Eliminating common terms and noting sin(½π-θ)
= cosθ,
this becomes-cosθ/2 +
2sinθ = 0,or
sinθ/cosθ = 1/4.
Using the identity,
tanθ = sinθ/cosθ,
we thus have θ = tan-1(1/4) = 14.0°.
The long side makes a 14.0° angle with the vertical.
- A uniform 400 N boom is supported as shown in
the figure below. Find the tension in the tie rope and the force
exerted on the boon by the pin at P.The problem mentions forces and looking at the diagram shows
that the object would rotate in the absence of any one of these
forces. This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the boom are a normal
from
the pin, the weight which acts from the centre of mass, and the
two tensions. We are given T2. The CM is obviously
at the centre of the boom. We do not have a simple method of
figuring out which way the normal points, instead we consider
it as two forces one vertical and one horizontal.ΣFx = 0 ΣFy = 0 Px – T1 = 0 Py – mg – T2 = 0 These tell us the obvious, that Px = T1
and Py = mg + T2 = 2400 N.We will use Method A for the torques since that method is
easiest
to apply here since the distances and angles are relatively easy
to find. We will take the pin as the pivot point since this eliminates
the torques from the pin.r F θ direction τz = rFsinθ 0 Px – – 0 0 Py – – 0 L/2 W 40° CW -LWsin(40°)/2 (3/4)L T1 50° CCW 3LT1sin(50°)/4 L T2 40° CW -LT2sin(40°) Since Στz = 0, the equation we get is
-WLsin(40°)/2 + 3LT1sin(50°)/4 –
LT2sin(40°) = 0 .Eliminating L from the above and rearranging to get T1
by itself yields,T1 = {2[W + 2T2]sin(40°)} /
3sin(50°) .Using the values we are given, we find T1 = 2461 N.
Since we know Px = T1, we also know the
pin
force isP = i2461 N + j2400
N.The magnitude of this force is P = [(Px)2
+ (Py)2 ]½ = 3438 N. The
force is directed at an angle θ = tan-1(Py/Px)
= 44.3° to the horizontal. Note that the pin force is not pointed
solely along the length of the boom as one might expect.Big Point To Remember: Pin forces are not always directed
in the obvious direction.
- In the figure below, a mass of 500 kg is held
motionless in the air by a 120-kg boom and a rope. Find the tension
in the rope. Find the force exerted on the boom by the pin at
P. The angles are θ
= 30.0°
and φ =
45.0°.The problem mentions forces and looking at the diagram
shows that the object would rotate in the absence of any one of
these forces. This indicates that we are dealing with a Static
Equilibrium problem. We solve Static Equilibrium problems by
sketching the extended free-body diagram, an FBD where the location
of the all forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the boom are
a normal from the pin, the weight which acts from the centre of
mass, and the two tensions. The CM is obviously at the centre
of the boom. We do not have a simple method of figuring out which
way the normal points, instead we consider it as two forces one
vertical and one horizontal.Fx = 0 ΣFy = 0 Px – T1cos(π/2-θ)
= 0Py – mg – T2 – T1sinθ
= 0These tell us that Px = T1cosθ
and Py = mg + T2
+ T1sinθ.
We are given the mass
of the load so we know T2 = mloadg = 4905
N.We will use Method A for the torques since that method is
easiest
to apply here since the distances and angles are relatively easy
to find. We will take the pin as the pivot point since this eliminates
the torques from the pin.r F θ direction τz = rFsinθ 0 Px – – 0 0 Py – – 0 L/2 mg π/2 – φ CW -Lmgsin(π/2-φ)/2 L T1 θ – φ CCW LT1sin(θ-φ) L T2 π/2 – φ CW -LT2sin(π/2-φ) Since Στz = 0, the equation we get is
-Lmg[sin(π/2-φ)]/2 + LT1sin(θ-φ)
– LT2sin(π/2-φ)
= 0 .Eliminating L from the above, multiplying through by 2, and using
the identity that sin(π/2-φ)
= cosφ yields,-mgcosφ
+ 2T1sin(θ-φ) –
2T2cosφ = 0.We rearrange to get T1 by itself,
T1 = (mg + 2T2)cosθ
/ 2sin(θ-φ) .Using the values we are given, and the value for T2,
we find T1 = 15008 N.We have Px = T1cosθ =
12998 N. As well,
Py = mg + T2 + T1sinθ
= 13587 N. Thus we also know that the pin force isP = i12998 N + j13587
N.The magnitude of this force is P = [(Px)2
+ (Py)2 ]½ = 1.88 × 104
N. The force is directed at an angle θ = tan-1(Py/Px)
= 46.3° to the horizontal. Note that the pin force is not pointed
solely along the length of the boom as one might expect.
- A rectangular sign of mass 50.0 kg and width
w = 5.00 m and l =
length 4.00 m is hanging from
a hinge and a rope as shown in the figure below. The rope makes and
angle
θ = 65.0°
with the right wall.
(a) Find the tension in the rope.
(b) Find the horizontal and vertical components of the hinge force.The problem mentions forces and looking at the diagram
shows that the object would rotate in the absence of any one of
these forces. This indicates that we are dealing with a Static
Equilibrium problem. We solve Static Equilibrium problems by
sketching the extended free-body diagram, an FBD where the location
of the all forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the sign are
a normal from the pin, the weight which acts from the centre of
mass, and the tension. The CM is obviously at the centre of the
sign. We do not have a simple method of figuring out which way
the normal points, instead we consider it as two forces one vertical
and one horizontal.ΣFx = 0 ΣFy = 0 Px + Tsinθ = 0 Py – mg + Tcosθ = 0 These tell us that Px = -Tsinθ and
Py = mg – Tcosθ.We will use Method B for the torques since that method is
easiest
to apply here since the location of the forces easy to find.
We will take the pin as the pivot point since this eliminates
the torques from the pin.x y Fx Fy τz = xFy – yFx 0 0 Px Py 0 w 0 Tsinθ Tcosθ wTcosθ w/2 -l/2 0 -mg –wmg/2 Since Στz = 0, the equation we get is
wTcosθ – wmg/2 = 0 .
Eliminating w from the above and rearranging yields,
T = mg / 2cosθ = 580.3 N .
The force equations give Px = -Tsinθ = -526 N and
Py = mg – Tcosτq = 245.0 N. The minus
sign for Px indicates we were wrong in assuming the the
hinge pushed the sign
to the right, it actually pulls the sign to the left.Thus the tension in the rope is 580 N and the horizontal and
vertical
components of the pin force are 526 N and 245 N respectively.
- Find the centre of mass of the object shown below.
Determine the tension in the strings and the unknown angle θ.
Each square has a side of length 32.0 cm. The object has a mass
of 125 g.To find the CM, we consider the squares as each
having mass M/3 located at their geometric centres. We will set
the origin at the upper right corner where the string is attached. Note
the symmetry of the object is such that the CM must be located
along a vertical line through the centre, that is the CM must
be located in the y axis and that xcm
= -0.16 m and zcm = 0.Piece Mass yi (m) miyi top M/3 -0.16 -0.053333M left M/3 -0.48 -0.16M right M/3 -0.48 -0.16M Totals: M -0.373333M Thus ycm = (Σmiyi)/Mtotal
= -0.373333 m.The problem mentions forces and looking at the diagram shows
that
the object would rotate in the absence of any one of these forces. This
indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the sign are the two
tensions
and the weight which acts from the centre of mass.ΣFx = 0 ΣFy = 0 T1sinθ – T2sin(65°)
= 0T1cosτq + T2cos(65°)
– Mg = 0These tell us that T1sinθ
= T2sin(65) and
T1cosθ + T2cos(65°) = Mg.We will use Method B for the torques since that method is
easiest
to apply here since the location of the forces easy to find. We
will locate the pivot at the upper right corner because we have
two unknowns there.x(m) y( m) Fx Fy τz = xFy – yFx 0 0 T1sinθ T1cosθ 0 -0.64 -0.32 -T2sin(65°) T2cos(65°) 0.64T2(65°)
– 0.32T2sin(65°)-0.16 -0.37333 0 -Mg 0.16Mg Since Στz = 0, the equation we get is
-0.64T2cos(65°)- 0.32T2sin(65°)
+ 0.16Mg = 0 .Rearranging the above yields the tension in the left string,
T2 = 0.16Mg / [0.64cos(65°) +
0.32sin(65°)]
= 0.3500 N .The force equations give
T1sinθ
= T2sin(65°) = 0.31725 N, andT1cosτ
= Mg – T2cos(65°) = 1.07831 N.Taking the ratio of these two results we have
sinθ/cosθ = 0.31725/1.07831
or tanθ = 0.2942. So the unknown angle is
θ = 16.4°. Substituting
the angle back into either of the two equations yields the tension
in the right string T1 = 1.124 N.
- The sign has a mass of 20.0 kg. The hinge is
located at the bottom of the left side. Find the centre of mass.
Determine the tension in the rope and the horizontal and vertical
components of the hinge force. The length, l,
is 12 cm.To find the CM, we break the sign into two pieces
each having all its mass located at their geometric centres.
Since the sign is uniform, its mass is proportional to its area.
The total area of the sign is 9l2. The crosspiece
has an area of 5l2 while the area of the vertical
piece is 4l2. If the sign has mass M, the crosspiece
therefore has a mass of (5/9)M and the vertical piece a mass of
(4/9)M. We will set the origin at the hinge Note the symmetry
of the object is such that the CM must be located along a vertical
line through the centre, that is the CM must be located in the
y axis and that xcm = 2.5l and
zcm = 0.Piece Mass yi (m) miyi top (5/9)M ½l (5/18)Ml bottom (4/9)M -2l -(8/9)Ml Totals: M -(11/18)Ml Thus ycm = (Σmiyi)/Mtotal
= -(11/18)l.The problem mentions forces and looking at the diagram shows
that
the object would rotate in the absence of any one of these forces.
This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the sign are the tension,
and the weight which acts from the centre of mass, and the normal
force from the hinge. Since we do not know the direction of the
normal force, we show components.ΣFx = 0 ΣFy = 0 -Hx + Tsin(40°) = 0 Hy + Tcos(40°) – Mg = 0 These tell us that Hx = Tsin(40°) and Hy
+ Tcos(40°) = Mg.We will use Method B for the torques since that method is
easiest
to apply here since the location of the forces easy to find. We
will locate the pivot at the hinge because we have two unknowns
there.x y Fx Fy τz = xFy – yFx 0 0 Hx Hy 0 5l l Tsin(40°) Tcos(40°) 5lTcos(40°) – lTsin(40°) (5/2)l (-11/18)l 0 -Mg -(5/2)lMg Since Στz = 0, the equation we get is
5lTcos(40°) – lTsin(40°) – (5/2)lMg
= 0 .Eliminating l and rearranging the above yields the tension
in the rope,T = (5/2)Mg / [5cos(40°) – sin(40°)] = 153.9 N .
The force equations give
Hx = Tsin(40°) = 98.9 N , and
Hy = Mg – Tcos(40°) = 78.3 N.
- A ladder is propped against a wall making an
angle with the floor. The wall is frictionless but the coefficients
of friction for the floor are μs and
μk respectively.
Obtain an expression for the smallest that can be if the ladder
is not to slip. Recall that tanθ
= sinθ/cosθ.The problem mentions forces and looking at the diagram shows that
the object would rotate in the absence of any one of these forces.
This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the ladder are the weight
which acts from the centre of mass, the normal forces from the
wall and floor, and friction. Since the ladder is not moving,
we are dealing with static friction. Since we want the smallest
angle, we are dealing with fs MAX. Since
the ladder has a tendency to move to the left, friction points
to the left. ΣFx = 0 ΣFy = 0 fs MAX – Nw =
0Nf – mg = 0 These tell us that fs MAX = μNw
and Nf = mg. We also know that fs MAX =
μsNf
where Nf is the normal force between the ladder and
floor. As a result, we have fs MAX = μsmg.
Hence Nw = μsmg as well.We will use Method A for the torques since that method is
easiest
to apply here since the distances and angles are easy to find.
We will locate the pivot at the floor because we have two unknowns
there.r (m) F (N) θ direction τz = rFsinθ 0 fs MAX – – 0 0 Nf – – 0 ½L mg π/2-θ CW -½Lmgsin(π/2-θ) L Nw θ CCW LNwsinθ Since τz = 0, the equation we get is
-½Lmgsin(π/2-θ)
+ LNwsinθ = 0 .Eliminating L and noting that sin(π/2-θ) = cosθ yields,
-½mgcosθ + Nwsinθ = 0 .
The force equations gave Nw = μsmg, so we
have-½mgcosθ
+ μsmgsinθ = 0 .Rearranging and using tanθ
= sinθ /cosτ, we getθ = tan-1(1 / 2μs) .
If the angle were any smaller than this, the ladder would slip.
- A truss is made by hinging two 3.0-m long uniform
planks, each of weight 150 N, as shown below. They rest on a
frictionless
floor and are kept from collapsing by a tie rope. A 500 N load
is held at the apex. Find the tension in the string. Hint – use
symmetry to solve the problem.This problem is impossible to solve without making use of symmetry.
That is the right and left planks are reflections of one another:
to solve the problem, we need only consider one plank. However
doing this means that we need to consider the force that one plank
exerts on the other. It is a normal force, and by Newton’s Third
Law, must be equal but opposite on each. This requires the normal
force to be horizontal as shown in the FBD of the left plank below.
Also note that each plank supports half the load since they are
identical.The problem mentions forces and looking at the diagram shows that
the object would collapse in the absence of any one of these forces.
This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the plank are the weight
which acts from the centre of mass, the normal forces from the
wall and other plank, the load, and tension.ΣFx = 0 ΣFy = 0 T – Nplank = 0 Nf – W – ½Wload
= 0These tell us that T = Nplank and Nf =
W + ½Wload = 400 N. A little trigonometry tells
us that θ = cos-1(1.75 / 3.00)
= 54.315°.We will use Method A for the torques since that method is
easiest
to apply here since the distance and angles are easy to find.
We will locate the pivot at the top of the plank because we have
two unknowns there.r (m) F (N) θ direction τz = rFsinθ 0 Nplank – – 0 0 ½Wload – – 0 3 Nf π/2-θ CW -3Nfsin(π/2-θ) 2.5 T θ CCW 2.5Tsinθ 1.5 W π/2+θ CCW 1.5mgsin(π/2+θ) Since Στz = 0, the equation we get is
-3Nfsin(π/2-θ)
+ 2.5Tsinθ
+ 1.5Wsin(π/2+θ) = 0 .We know that sin(π/2-θ)
= sin(π/2+θ)
= cosθ and we already determined
that Nf = W + ½Wload, so our torque
equation becomes-3[W + ½Wload]cosθ
+ 2.5Tsinθ + 1.5Wcosθ
= 0 .We can rearrange this to find T
T = {3[W + ½Wload]cosθ
– 1.5Wcosθ }/2.5sinθ
= 3[W + Wload]/ 5tanθ = 280.1 N.This is also the value of Nplank, the normal force
from one plank to the other.
- A wheel of mass M and radius R rests on a horizontal
surface against a step of height h (h < R). A horizontal force
F applied to the axle of the wheel just raises the wheel off the
step. Find the force F.
The problem mentions forces and looking at the diagram shows that
the object would roll or rotate. This indicates that we are dealing
with a Static Equilibrium problem. We solve Static Equilibrium
problems by sketching the extended free-body diagram, an FBD where
the location of the all forces are indicated so that torques can
be calculated. Then we determine the three equations necessary
for static equilibrium, ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the plank are the weight
which acts from the centre of mass, the normal forces from the
wall and from the step, and the applied force. We do not know
the direction of the normal force from the step, so we will consider
it horizontal and vertical components.ΣFx = 0 ΣFy = 0 F – Nx = 0 Nf + Ny – Mg = 0 These tell us that F = Nx and Nf + Ny
= Mg. Also recall that if the wheel leaves the ground, Nf
= 0 and thus Ny = Mg.We will use Method B for the torques since that method is
easiest
to apply here since the location of each force can be found with
the help of some geometry. We will locate the pivot at the top
of the step because we have two unknowns there. The y
locations of the forces, F, Mg, and Nf are easy to
read from the diagram. The x location is the same
for each but takes a little work as shown in the diagram below
where it can be seen that x = [R2 – (R-h)2]½.x y Fx Fy τz = xFy – yFx 0 0 Nx Ny 0 -[R2 – (R-h)2]½ R-h 0 -Mg [R2 – (R-h)2]½Mg -[R2 – (R-h)2]½ R-h F 0 -(R-h)F -[R2 – (R-h)2]½ -h 0 Nf -[R2 – (R-h)2]½Nf Since Στz = 0, the equation we get is
[R2 – (R-h)2]½Mg
– (R-h)F – [R2 – (R-h)2]½Nf
= 0 .As pointed out, the wheel just loses contact with the ground
when Nf = 0. That gives us our expression for F,F = Mg {[R2 – (R-h)2]½
/ (R-h)} .For any applied force less than this value, the wheel remains
in contact with the ground.
Physics 1120: Centre of Mass & Static Equilibrium Solutions
Questions: 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Centre of Mass
- The distance between the oxygen molecule and
each of the hydrogen
atoms in a water (H2O) molecule is 0.0958 nm; the angle
between the two oxygen-hydrogen bonds is 105°.
Treating the atoms as particles, find the centre of mass.The problem expects you to recall that the mass
of an oxygen atom is 16 times that of a hydrogen atom. The first
step is to choose a coordinate system, such as the one in the
diagram, and locate each particle. The chosen origin is the centre
of the box.Atom Mass (H) xi yi mixi miyi H 1 -0.0958sin15 0.0958cos15 -0.02479 0.09254 O 16 0 0 0 0 H 1 0.0958 0 0.0958 0 Totals: 18 0.07101 0.09254 The coordinates of the centre of mass are given by
xcm = (Σmixi)/Mtotal
= 0.07101/18 = 0.0039 nm,and
ycm = (miyi)/Mtotal
= 0.09254/18 = 0.0051 nm.The centre of mass is located at (0.0039 nm, 0.0051 nm). Answers
will vary based on the choice of coordinate system.
- Where is the centre of mass of a uniform cubic box of side
length L which has no lid?In dealing with real objects rather than particles, we treat
the complex object as a grouping of simpler shapes. The CM of
the simpler shapes is at their easy to find geometric centre if
the object is uniform. Each pierce can now be considered a particle
with the mass of the piece located at the CM of that piece. We
have, in effect, turned the complex shape into a collection of
particles. In this case, each of the five sides can be considered
a separate particle.The next step is to choose a coordinate system, such as the one
in the diagram below, and locate each particle. The origin is
at the centre of the box.Side Mass xi yi zi mixi miyi mizi bottom M 0 0 -½L 0 0 -½ML front M 0 -½L 0 0 -½ML 0 back M 0 ½L 0 0 ½ML 0 left M -½L 0 0 -½ML 0 0 right M ½L 0 0 ½ML 0 0 Totals: 5M 0 0 -½ML The coordinates of the centre of mass are given by
xcm = (Σmixi)/Mtotal
= 0,ycm = (Σmiyi)/Mtotal
= 0,and
zcm = (Σmizi)/Mtotal
= -½ML / 5M = -L/10.The centre of mass is located at (0, 0, -L/10). Answers will
vary based on the choice of coordinate system.It is also permissible to use symmetry arguments. For example,
the figure in the diagram is only unbalanced in the z direction,
thus we know that xcm = ycm = 0. We only
needed the z columns in the above table.
- Two uniform squares of sheet metal of dimension L ×
L are joined at a right angle along one edge. One of the squares
has twice the mass of the other. Find the centre of mass.In dealing with real objects rather than particles,
we treat the complex object as a grouping of simpler shapes.
The CM of the simpler shapes is at their easy to find geometric
centre if the object is uniform. Each pierce can now be considered
a particle with the mass of the piece located at the CM of that
piece. We have, in effect, turned the complex shape into a collection
of particles. In this case, we have one particle of mass M located
in the centre of the lighter side, and a mass of 2M in the centre
of the heavier side.The next step is to choose a coordinate system,
such as the one in the diagram below, and locate each particle. The
origin is in the centre of the join of the two plates.Using the symmetry of the problem, we see that the
CM must be located in the xz plane, we know that
ycm = 0.Side Mass xi zi mixi mizi side M 0 ½L 0 ½ML bottom 2M ½L 0 ML 0 Totals: 3M ML ½ML The components of the centre of mass are given by
xcm = (Σmixi)/Mtotal
= ML / 3M = L/3,zcm = (Σmizi)/Mtotal
= ½ML / 3M = L/6.The centre of mass is located at (L/3, 0, L/6). Answers will
vary based on the choice of coordinate system.
- A cube of iron has dimension L ×
L × L. A hole of radius ¼L
has been drilled all the way through the cube, so that one side
of the hole is tangent to one face along its entire length. Where
is the centre of mass of the drilled cube?In dealing with real objects rather than particles,
we treat the complex object as a grouping of simpler shapes.
The CM of the simpler shapes is at their easy to find geometric
centre if the object is uniform. Each pierce can now be considered
a particle with the mass of the piece located at the CM of that
piece. We have, in effect, turned the complex shape into a collection
of particles. In this case, we have a solid cube and a cylindrical
hole. We treat holes as objects of negative mass.To proceed we need to know the mass of the cylindrical
hole. Since the object was uniform, its mass is proportional
to its volume. The solid cube had a mass M and a volume L3.
The cylinder has a volume Vcyl = πr2L =
πL3/16. Thus the mass of the cylindrical hole ismcyl = mcube[Vcyl
/ Vcube] = M[(πL3/16) / L3]
= πM/16.The next step is to choose a coordinate system,
such as the one in the diagram below, and locate each particle.Using the symmetry of the problem, we see that the
CM must be located in the x axis, we know that ycm
= zcm = 0.Side Mass xi mixi solid cube M 0 0 hole -πM/16 -¼L πML/64 Totals: M(1-π/16) πML/64 The location of the x component of the centre of mass is given
byxcm = (Σmixi)/Mtotal
= -πML/64 / -M(1-π/16)
= πL / 64(1-π/16).The centre of mass is located at (πL / 64(1-π/16), 0, 0).
Answers will vary based on the choice of coordinate system.
- An 80-kg logger is standing on one end of a 10-m long, 300-kg,
tree trunk in the middle of the Fraser River. The logger walks
upriver along the trunk to the other end of the log. As a result
the log moves some distance L down river. What is the displacement
L?The logger and the log are a system; the system has a certain
centre of mass. The motion of the logger is an internal force;
internal forces cannot change the centre of mass of the system.Examining the diagram, the log has moved down the river a distance
equal to twice the distance between the centre of the log and
the CM of the logger-log system. We need to find the CM of the
logger-log system. Taking the origin as the end of the log,xcm = (Σmixi)/Mtotal
= [80×0 + 300×5] / 380 = 3.047 m.Since the centre of the log is at 5 m, distance between the CM
and the centre of the log is 1.05 m. The log moved twice this
distance or 2.11 m.
- A shell is fired at 25 m/s at 25 above the horizontal. At
the top of its parabolic flight, it breaks into two pieces. One
piece, having two-thirds of the total mass of the shell lands
60 m from where the shell was fired. Where did the other piece
land?The pieces of the shell are a system; the system has a certain
centre of mass. The explosion is an internal force; internal
forces cannot change the centre of mass of the system. The CM
of the pieces will land where the CM of an unexploded shell will
land.The first step is to find xcm, the landing position
of the shell. That involves solving the projectile motion problem.i j x = xcm = ? y = 0 ax = 0 ay = -g = -9.81 m/s v0x = 25cos25 = 22.658 m/s v0y = 25sin25 = 10.565 m/s t = ? t = ? The j information allows us to find the time in
air using y = v0yt – ½gt2. Since y
= 0, this reduces tot = 2voy/g = 2×10.565 / 9.81 = 2.154 s.
We then find the landing position using x = v0xt +
½axt2. Since ax = 0,xcm = voxt = 22.658 × 2.154 =
48.806 m.The centre of mass is determined by the formula
xcm = (Σmixi)/Mtotal
= m1x1/Mtotal + m2x2/Mtotal.Since we now know xcm and x2, we can rearrange
to find m1,x1 = [Mtotalxcm – m2x2]/m1
= [1×48.806 – (2/3)60] / (1/3) = 26.4 m.So the smaller piece lands 26.4 m from where the shell was fired.
Torque and Static Equilibrium
- In the diagram below, three forces are applied
to a 3-4-5 triangle. The forces are F1 = 91.7 N, F2
= 150 N, and F3 = 67.7 N. F3 is applied
at the middle of side AB. (a) Find the net torque about point
A where F1 acts. (b) Find the net torque about point B at the lower right corner of the triangle. (c) Find the net torque
about point C where F2 acts.CThere are two methods for determining torque. Method A is to
use τz
= rFsinθ, where r is the distance from the pivot
point to the point where the force F acts and
θ is the angle between
r and F. The sign of τz is found
using
the right-hand rule. Method B is to use τz = xFy
– yFx, where (x, y) is the location of where the force
is acting taken relative to the pivot point which is taken to
be the origin (0, 0). Fx and Fy are the
components of the force – careful attention must be paid to signs.Method A.
First note that the interior angles of the triangle are α
= tan-1(4/3) = 53.130°, and γ
= tan-1 (3/4) = 36.870°. F2 makes an angle
φ = 180° – 110°
– γ = 16.870° with the vertical.(a)
r (m) F (N) θ direction τz = rFsinθ (N-m) 0 91.7 – – 0 5 150 110° CCW 704.769 2 67.7 130° CW -103.722 Total: 601.0 (b)
r (m) F (N) θ direction τz = rFsinθ (N-m) 4 91.7 90° CCW 366.800 3 150 110° + γ CCW 130.591 2 67.7 50° CCW 103.722 Total: 601.1 (c)
r (m) F (N) direction τz = rFsinθ (N-m) 5 91.7 90° + α CCW 366.800 0 150 – – 0 3.60555 67.7 50° + 56.130° CCW 234.272 Total: 601.1 Method B:
(a)
x (m) y (m) Fx (N) Fy (N) τz = xFy – yFx
(N-m)0 0 0 -91.7 0 4 3 -150sinφ 150cosφ 704.769 2 0 67.7cos50° -67.7sin50° -103.722 total: 601.0 (b)
x (m) y (m) Fx (N) Fy (N) τz = xFy – yFx
(N-m)-4 0 0 -91.7 103.722 0 3 -150sinφ 150cosφ 130.591 -2 0 67.7cos50° -67.7sin50° 366.80 total: 601.1 (c)
x (m) y (m) Fx (N) Fy (N) τz = xFy – yFx
(N-m)-4 -3 0 -91.7 366.800 0 0 -150sinφ 150cosφ 0 -2 -3 67.7cos50° -67.7sin50° 234.272 total: 601.1 Please note that the only reason the total torque at point A,
B, and C are the same is because F1 + F2
+ F3 = 0.
- An L-shaped object of uniform density is hung
over a nail so that it is free to pivot. What angle, θ,
does the long side make with the vertical? The long side of the
L-shaped object is twice as long as the short side?The problem mentioned that the object is free to
pivot, to rotate. This indicates that we are dealing with a Static
Equilibrium problem. We solve Static Equilibrium problems by
sketching the extended free-body diagram, an FBD where the location
of the all forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the L-shaped
object are a normal from the nail and the weight which acts from
the centre of mass. Ordinarily, for complex shapes, we first
determine the CM. However, in this case, it is easier to consider
the two arms of the objects as being separate objects. The long
arm will have a mass (2/3)mg and the short arm will be (1/3)mg.
We do not have a simple method of figuring out which way the
normal points. As with all pins, we consider it as two forces
one vertical and one horizontal.ΣFx = 0 ΣFy = 0 Nx = 0 Ny – (1/3)mg – (2/3)mg = 0 These tell us the obvious, the normal has no horizontal component
and that it supports the weight of the object.We will use Method A for the torques since that method is
easiest
to apply here. We will take the nail as the pivot point since
this eliminates the torques from the nail.r F direction τz = rFsinθ 0 Nx – – 0 0 Ny – – 0 L/2 (1/3)mg ½π-θ CCW mgLsin(½π-θ)/6 L (2/3)mg θ CW −2mgLsinθ/3 Since Στz = 0, the equation we get is
mgLsin(½π-θ)/6
+ −2mgLsinθ/3 = 0.Eliminating common terms and noting sin(½π-θ)
= cosθ,
this becomes-cosθ/2 +
2sinθ = 0,or
sinθ/cosθ = 1/4.
Using the identity,
tanθ = sinθ/cosθ,
we thus have θ = tan-1(1/4) = 14.0°.
The long side makes a 14.0° angle with the vertical.
- A uniform 400 N boom is supported as shown in
the figure below. Find the tension in the tie rope and the force
exerted on the boon by the pin at P.The problem mentions forces and looking at the diagram shows
that the object would rotate in the absence of any one of these
forces. This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the boom are a normal
from
the pin, the weight which acts from the centre of mass, and the
two tensions. We are given T2. The CM is obviously
at the centre of the boom. We do not have a simple method of
figuring out which way the normal points, instead we consider
it as two forces one vertical and one horizontal.ΣFx = 0 ΣFy = 0 Px – T1 = 0 Py – mg – T2 = 0 These tell us the obvious, that Px = T1
and Py = mg + T2 = 2400 N.We will use Method A for the torques since that method is
easiest
to apply here since the distances and angles are relatively easy
to find. We will take the pin as the pivot point since this eliminates
the torques from the pin.r F θ direction τz = rFsinθ 0 Px – – 0 0 Py – – 0 L/2 W 40° CW -LWsin(40°)/2 (3/4)L T1 50° CCW 3LT1sin(50°)/4 L T2 40° CW -LT2sin(40°) Since Στz = 0, the equation we get is
-WLsin(40°)/2 + 3LT1sin(50°)/4 –
LT2sin(40°) = 0 .Eliminating L from the above and rearranging to get T1
by itself yields,T1 = {2[W + 2T2]sin(40°)} /
3sin(50°) .Using the values we are given, we find T1 = 2461 N.
Since we know Px = T1, we also know the
pin
force isP = i2461 N + j2400
N.The magnitude of this force is P = [(Px)2
+ (Py)2 ]½ = 3438 N. The
force is directed at an angle θ = tan-1(Py/Px)
= 44.3° to the horizontal. Note that the pin force is not pointed
solely along the length of the boom as one might expect.Big Point To Remember: Pin forces are not always directed
in the obvious direction.
- In the figure below, a mass of 500 kg is held
motionless in the air by a 120-kg boom and a rope. Find the tension
in the rope. Find the force exerted on the boom by the pin at
P. The angles are θ
= 30.0°
and φ =
45.0°.The problem mentions forces and looking at the diagram
shows that the object would rotate in the absence of any one of
these forces. This indicates that we are dealing with a Static
Equilibrium problem. We solve Static Equilibrium problems by
sketching the extended free-body diagram, an FBD where the location
of the all forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the boom are
a normal from the pin, the weight which acts from the centre of
mass, and the two tensions. The CM is obviously at the centre
of the boom. We do not have a simple method of figuring out which
way the normal points, instead we consider it as two forces one
vertical and one horizontal.Fx = 0 ΣFy = 0 Px – T1cos(π/2-θ)
= 0Py – mg – T2 – T1sinθ
= 0These tell us that Px = T1cosθ
and Py = mg + T2
+ T1sinθ.
We are given the mass
of the load so we know T2 = mloadg = 4905
N.We will use Method A for the torques since that method is
easiest
to apply here since the distances and angles are relatively easy
to find. We will take the pin as the pivot point since this eliminates
the torques from the pin.r F θ direction τz = rFsinθ 0 Px – – 0 0 Py – – 0 L/2 mg π/2 – φ CW -Lmgsin(π/2-φ)/2 L T1 θ – φ CCW LT1sin(θ-φ) L T2 π/2 – φ CW -LT2sin(π/2-φ) Since Στz = 0, the equation we get is
-Lmg[sin(π/2-φ)]/2 + LT1sin(θ-φ)
– LT2sin(π/2-φ)
= 0 .Eliminating L from the above, multiplying through by 2, and using
the identity that sin(π/2-φ)
= cosφ yields,-mgcosφ
+ 2T1sin(θ-φ) –
2T2cosφ = 0.We rearrange to get T1 by itself,
T1 = (mg + 2T2)cosθ
/ 2sin(θ-φ) .Using the values we are given, and the value for T2,
we find T1 = 15008 N.We have Px = T1cosθ =
12998 N. As well,
Py = mg + T2 + T1sinθ
= 13587 N. Thus we also know that the pin force isP = i12998 N + j13587
N.The magnitude of this force is P = [(Px)2
+ (Py)2 ]½ = 1.88 × 104
N. The force is directed at an angle θ = tan-1(Py/Px)
= 46.3° to the horizontal. Note that the pin force is not pointed
solely along the length of the boom as one might expect.
- A rectangular sign of mass 50.0 kg and width
w = 5.00 m and l =
length 4.00 m is hanging from
a hinge and a rope as shown in the figure below. The rope makes and
angle
θ = 65.0°
with the right wall.
(a) Find the tension in the rope.
(b) Find the horizontal and vertical components of the hinge force.The problem mentions forces and looking at the diagram
shows that the object would rotate in the absence of any one of
these forces. This indicates that we are dealing with a Static
Equilibrium problem. We solve Static Equilibrium problems by
sketching the extended free-body diagram, an FBD where the location
of the all forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the sign are
a normal from the pin, the weight which acts from the centre of
mass, and the tension. The CM is obviously at the centre of the
sign. We do not have a simple method of figuring out which way
the normal points, instead we consider it as two forces one vertical
and one horizontal.ΣFx = 0 ΣFy = 0 Px + Tsinθ = 0 Py – mg + Tcosθ = 0 These tell us that Px = -Tsinθ and
Py = mg – Tcosθ.We will use Method B for the torques since that method is
easiest
to apply here since the location of the forces easy to find.
We will take the pin as the pivot point since this eliminates
the torques from the pin.x y Fx Fy τz = xFy – yFx 0 0 Px Py 0 w 0 Tsinθ Tcosθ wTcosθ w/2 -l/2 0 -mg –wmg/2 Since Στz = 0, the equation we get is
wTcosθ – wmg/2 = 0 .
Eliminating w from the above and rearranging yields,
T = mg / 2cosθ = 580.3 N .
The force equations give Px = -Tsinθ = -526 N and
Py = mg – Tcosτq = 245.0 N. The minus
sign for Px indicates we were wrong in assuming the the
hinge pushed the sign
to the right, it actually pulls the sign to the left.Thus the tension in the rope is 580 N and the horizontal and
vertical
components of the pin force are 526 N and 245 N respectively.
- Find the centre of mass of the object shown below.
Determine the tension in the strings and the unknown angle θ.
Each square has a side of length 32.0 cm. The object has a mass
of 125 g.To find the CM, we consider the squares as each
having mass M/3 located at their geometric centres. We will set
the origin at the upper right corner where the string is attached. Note
the symmetry of the object is such that the CM must be located
along a vertical line through the centre, that is the CM must
be located in the y axis and that xcm
= -0.16 m and zcm = 0.Piece Mass yi (m) miyi top M/3 -0.16 -0.053333M left M/3 -0.48 -0.16M right M/3 -0.48 -0.16M Totals: M -0.373333M Thus ycm = (Σmiyi)/Mtotal
= -0.373333 m.The problem mentions forces and looking at the diagram shows
that
the object would rotate in the absence of any one of these forces. This
indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the sign are the two
tensions
and the weight which acts from the centre of mass.ΣFx = 0 ΣFy = 0 T1sinθ – T2sin(65°)
= 0T1cosτq + T2cos(65°)
– Mg = 0These tell us that T1sinθ
= T2sin(65) and
T1cosθ + T2cos(65°) = Mg.We will use Method B for the torques since that method is
easiest
to apply here since the location of the forces easy to find. We
will locate the pivot at the upper right corner because we have
two unknowns there.x(m) y( m) Fx Fy τz = xFy – yFx 0 0 T1sinθ T1cosθ 0 -0.64 -0.32 -T2sin(65°) T2cos(65°) 0.64T2(65°)
– 0.32T2sin(65°)-0.16 -0.37333 0 -Mg 0.16Mg Since Στz = 0, the equation we get is
-0.64T2cos(65°)- 0.32T2sin(65°)
+ 0.16Mg = 0 .Rearranging the above yields the tension in the left string,
T2 = 0.16Mg / [0.64cos(65°) +
0.32sin(65°)]
= 0.3500 N .The force equations give
T1sinθ
= T2sin(65°) = 0.31725 N, andT1cosτ
= Mg – T2cos(65°) = 1.07831 N.Taking the ratio of these two results we have
sinθ/cosθ = 0.31725/1.07831
or tanθ = 0.2942. So the unknown angle is
θ = 16.4°. Substituting
the angle back into either of the two equations yields the tension
in the right string T1 = 1.124 N.
- The sign has a mass of 20.0 kg. The hinge is
located at the bottom of the left side. Find the centre of mass.
Determine the tension in the rope and the horizontal and vertical
components of the hinge force. The length, l,
is 12 cm.To find the CM, we break the sign into two pieces
each having all its mass located at their geometric centres.
Since the sign is uniform, its mass is proportional to its area.
The total area of the sign is 9l2. The crosspiece
has an area of 5l2 while the area of the vertical
piece is 4l2. If the sign has mass M, the crosspiece
therefore has a mass of (5/9)M and the vertical piece a mass of
(4/9)M. We will set the origin at the hinge Note the symmetry
of the object is such that the CM must be located along a vertical
line through the centre, that is the CM must be located in the
y axis and that xcm = 2.5l and
zcm = 0.Piece Mass yi (m) miyi top (5/9)M ½l (5/18)Ml bottom (4/9)M -2l -(8/9)Ml Totals: M -(11/18)Ml Thus ycm = (Σmiyi)/Mtotal
= -(11/18)l.The problem mentions forces and looking at the diagram shows
that
the object would rotate in the absence of any one of these forces.
This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the sign are the tension,
and the weight which acts from the centre of mass, and the normal
force from the hinge. Since we do not know the direction of the
normal force, we show components.ΣFx = 0 ΣFy = 0 -Hx + Tsin(40°) = 0 Hy + Tcos(40°) – Mg = 0 These tell us that Hx = Tsin(40°) and Hy
+ Tcos(40°) = Mg.We will use Method B for the torques since that method is
easiest
to apply here since the location of the forces easy to find. We
will locate the pivot at the hinge because we have two unknowns
there.x y Fx Fy τz = xFy – yFx 0 0 Hx Hy 0 5l l Tsin(40°) Tcos(40°) 5lTcos(40°) – lTsin(40°) (5/2)l (-11/18)l 0 -Mg -(5/2)lMg Since Στz = 0, the equation we get is
5lTcos(40°) – lTsin(40°) – (5/2)lMg
= 0 .Eliminating l and rearranging the above yields the tension
in the rope,T = (5/2)Mg / [5cos(40°) – sin(40°)] = 153.9 N .
The force equations give
Hx = Tsin(40°) = 98.9 N , and
Hy = Mg – Tcos(40°) = 78.3 N.
- A ladder is propped against a wall making an
angle with the floor. The wall is frictionless but the coefficients
of friction for the floor are μs and
μk respectively.
Obtain an expression for the smallest that can be if the ladder
is not to slip. Recall that tanθ
= sinθ/cosθ.The problem mentions forces and looking at the diagram shows that
the object would rotate in the absence of any one of these forces.
This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the ladder are the weight
which acts from the centre of mass, the normal forces from the
wall and floor, and friction. Since the ladder is not moving,
we are dealing with static friction. Since we want the smallest
angle, we are dealing with fs MAX. Since
the ladder has a tendency to move to the left, friction points
to the left.ΣFx = 0 ΣFy = 0 fs MAX – Nw =
0Nf – mg = 0 These tell us that fs MAX = μNw
and Nf = mg. We also know that fs MAX =
μsNf
where Nf is the normal force between the ladder and
floor. As a result, we have fs MAX = μsmg.
Hence Nw = μsmg as well.We will use Method A for the torques since that method is
easiest
to apply here since the distances and angles are easy to find.
We will locate the pivot at the floor because we have two unknowns
there.r (m) F (N) θ direction τz = rFsinθ 0 fs MAX – – 0 0 Nf – – 0 ½L mg π/2-θ CW -½Lmgsin(π/2-θ) L Nw θ CCW LNwsinθ Since τz = 0, the equation we get is
-½Lmgsin(π/2-θ)
+ LNwsinθ = 0 .Eliminating L and noting that sin(π/2-θ) = cosθ yields,
-½mgcosθ + Nwsinθ = 0 .
The force equations gave Nw = μsmg, so we
have-½mgcosθ
+ μsmgsinθ = 0 .Rearranging and using tanθ
= sinθ /cosτ, we getθ = tan-1(1 / 2μs) .
If the angle were any smaller than this, the ladder would slip.
- A truss is made by hinging two 3.0-m long uniform
planks, each of weight 150 N, as shown below. They rest on a
frictionless
floor and are kept from collapsing by a tie rope. A 500 N load
is held at the apex. Find the tension in the string. Hint – use
symmetry to solve the problem.This problem is impossible to solve without making use of symmetry.
That is the right and left planks are reflections of one another:
to solve the problem, we need only consider one plank. However
doing this means that we need to consider the force that one plank
exerts on the other. It is a normal force, and by Newton’s Third
Law, must be equal but opposite on each. This requires the normal
force to be horizontal as shown in the FBD of the left plank below.
Also note that each plank supports half the load since they are
identical.The problem mentions forces and looking at the diagram shows that
the object would collapse in the absence of any one of these forces.
This indicates that we are dealing with a Static Equilibrium
problem. We solve Static Equilibrium problems by sketching the
extended free-body diagram, an FBD where the location of the all
forces are indicated so that torques can be calculated. Then
we determine the three equations necessary for static equilibrium,
ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the plank are the weight
which acts from the centre of mass, the normal forces from the
wall and other plank, the load, and tension.ΣFx = 0 ΣFy = 0 T – Nplank = 0 Nf – W – ½Wload
= 0These tell us that T = Nplank and Nf =
W + ½Wload = 400 N. A little trigonometry tells
us that θ = cos-1(1.75 / 3.00)
= 54.315°.We will use Method A for the torques since that method is
easiest
to apply here since the distance and angles are easy to find.
We will locate the pivot at the top of the plank because we have
two unknowns there.r (m) F (N) θ direction τz = rFsinθ 0 Nplank – – 0 0 ½Wload – – 0 3 Nf π/2-θ CW -3Nfsin(π/2-θ) 2.5 T θ CCW 2.5Tsinθ 1.5 W π/2+θ CCW 1.5mgsin(π/2+θ) Since Στz = 0, the equation we get is
-3Nfsin(π/2-θ)
+ 2.5Tsinθ
+ 1.5Wsin(π/2+θ) = 0 .We know that sin(π/2-θ)
= sin(π/2+θ)
= cosθ and we already determined
that Nf = W + ½Wload, so our torque
equation becomes-3[W + ½Wload]cosθ
+ 2.5Tsinθ + 1.5Wcosθ
= 0 .We can rearrange this to find T
T = {3[W + ½Wload]cosθ
– 1.5Wcosθ }/2.5sinθ
= 3[W + Wload]/ 5tanθ = 280.1 N.This is also the value of Nplank, the normal force
from one plank to the other.
- A wheel of mass M and radius R rests on a horizontal
surface against a step of height h (h < R). A horizontal force
F applied to the axle of the wheel just raises the wheel off the
step. Find the force F.The problem mentions forces and looking at the diagram shows that
the object would roll or rotate. This indicates that we are dealing
with a Static Equilibrium problem. We solve Static Equilibrium
problems by sketching the extended free-body diagram, an FBD where
the location of the all forces are indicated so that torques can
be calculated. Then we determine the three equations necessary
for static equilibrium, ΣFx = 0,
ΣFy = 0,
and Στz = 0.The forces that we know are working on the plank are the weight
which acts from the centre of mass, the normal forces from the
wall and from the step, and the applied force. We do not know
the direction of the normal force from the step, so we will consider
it horizontal and vertical components.ΣFx = 0 ΣFy = 0 F – Nx = 0 Nf + Ny – Mg = 0 These tell us that F = Nx and Nf + Ny
= Mg. Also recall that if the wheel leaves the ground, Nf
= 0 and thus Ny = Mg.We will use Method B for the torques since that method is
easiest
to apply here since the location of each force can be found with
the help of some geometry. We will locate the pivot at the top
of the step because we have two unknowns there. The y
locations of the forces, F, Mg, and Nf are easy to
read from the diagram. The x location is the same
for each but takes a little work as shown in the diagram below
where it can be seen that x = [R2 – (R-h)2]½.x y Fx Fy τz = xFy – yFx 0 0 Nx Ny 0 -[R2 – (R-h)2]½ R-h 0 -Mg [R2 – (R-h)2]½Mg -[R2 – (R-h)2]½ R-h F 0 -(R-h)F -[R2 – (R-h)2]½ -h 0 Nf -[R2 – (R-h)2]½Nf Since Στz = 0, the equation we get is
[R2 – (R-h)2]½Mg
– (R-h)F – [R2 – (R-h)2]½Nf
= 0 .As pointed out, the wheel just loses contact with the ground
when Nf = 0. That gives us our expression for F,F = Mg {[R2 – (R-h)2]½
/ (R-h)} .For any applied force less than this value, the wheel remains
in contact with the ground.