physics 1120: centre of mass & static equilibrium solutions of science... · questions:...

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Physics 1120: Centre of Mass & Static Equilibrium Solutions

Centre of Mass

1. The distance between the oxygen molecule and each of the hydrogen atoms in a water (H2O) molecule is0.0958 nm; the angle between the two oxygenhydrogen 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 miyiH 1 0.0958sin15 0.0958cos15 0.02479 0.09254O 16 0 0 0 0H 1 0.0958 0 0.0958 0Totals: 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 ofcoordinate system.

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2. 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 simplershapes. 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 fivesides 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 eachparticle. The origin is at the centre of the box.

Side Mass xi yi zi mixi miyi mizibottom M 0 0 ½L 0 0 ½MLfront M 0 ½L 0 0 ½ML 0back M 0 ½L 0 0 ½ML 0left M ½L 0 0 ½ML 0 0right M ½L 0 0 ½ML 0 0Totals: 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 coordinatesystem.

It is also permissible to use symmetry arguments. For example, the figure in the diagram is only unbalancedin the z direction, thus we know that xcm = ycm = 0. We only needed the z columns in the above table.

3. Two uniform squares of sheet metal of dimension L × L are joined at a right angle along one edge. One ofthe 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 simplershapes. 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 oneparticle of mass M located in the centre of the lighter side, and a mass of 2M in the centre of the heavierside.

The next step is to choose a coordinate system, such as the one in the diagram below, and locate eachparticle. 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 thatycm = 0.

Side Mass xi zi mixi miziside M 0 ½L 0 ½MLbottom 2M ½L 0 ML 0Totals: 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 coordinatesystem.

4. 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 massof the drilled cube?

In dealing with real objects rather than particles, we treat the complex object as a grouping of simplershapes. 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 solidcube 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 isproportional to its volume. The solid cube had a mass M and a volume L3. The cylinder has a volume Vcyl= r2L = L3/16. Thus the mass of the cylindrical hole is

mcyl = 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 eachparticle.

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 mixisolid cube M 0 0hole πM/16 ¼L πML/64Totals: M(1π/16) πML/64

The location of the x component of the centre of mass is given by

xcm = (Σmixi)/Mtotal = πML/64 / M(1π/16) = πL / 64(1π/16).

The centre of mass is located at (πL / 64(1π/16), 0, 0). Answers will vary based on the choice ofcoordinate system.

5. An 80kg logger is standing on one end of a 10m long, 300kg, tree trunk in the middle of the FraserRiver. The logger walks upriver along the trunk to the other end of the log. As a result the log moves somedistance 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 isan 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 betweenthe centre of the log and the CM of the loggerlog system. We need to find the CM of the loggerlogsystem. 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 logmoved twice this distance or 2.11 m.

6. A shell is fired at 25 m/s at 25 above the horizontal. At the top of its parabolic flight, it breaks into twopieces. One piece, having twothirds of the total mass of the shell lands 60 m from where the shell wasfired. 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 internalforce; internal forces cannot change the centre of mass of the system. The CM of the pieces will landwhere 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 motionproblem.

i jx = xcm = ? y = 0ax = 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 to

t = 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

7. In the diagram below, three forces are applied to a 345 triangle. The forces are F1 = 91.7 N, F2 = 150N, 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 distancefrom the pivot point to the point where the force F acts and θ is the angle between r and F. The sign of τzis found using the righthand rule. Method B is to use τz = xFy yFx, where (x, y) is the location of wherethe force is acting taken relative to the pivot point which is taken to be the origin (0, 0). Fx and Fy are thecomponents of the force careful attention must be paid to signs.

Method A.

First note that the interior angles of the triangle are α = tan1(4/3) = 53.130°, and γ = tan1 (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 (Nm)

0 91.7 05 150 110° CCW 704.7692 67.7 130° CW 103.722

Total: 601.0

(b)

r (m) F (N) θ direction τz = rFsinθ (Nm)

4 91.7 90° CCW 366.8003 150 110° + γ CCW 130.5912 67.7 50° CCW 103.722

Total: 601.1

(c)

r (m) F (N) direction τz = rFsinθ (Nm)

5 91.7 90° + α CCW 366.8000 150 03.60555 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 (Nm)

0 0 0 91.7 04 3 150sinφ 150cosφ 704.7692 0 67.7cos50° 67.7sin50° 103.722

total: 601.0

(b)


4 0 0 91.7 103.7220 3 150sinφ 150cosφ 130.5912 0 67.7cos50° 67.7sin50° 366.80

total: 601.1

(c)


4 3 0 91.7 366.8000 0 150sinφ 150cosφ 02 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.

8. An Lshaped object of uniform density is hung over a nail so that it is free to pivot. What angle, θ, doesthe long side make with the vertical? The long side of the Lshaped object is twice as long as the shortside?

The problem mentioned that the object is free to pivot, to rotate. This indicates that we are dealing with aStatic Equilibrium problem. We solve Static Equilibrium problems by sketching the extended freebodydiagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Thenwe determine the three equations necessary for static equilibrium, ΣFx = 0, ΣFy = 0, and Στz = 0.

The forces that we know are working on the Lshaped object are a normal from the nail and the weightwhich 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 armwill have a mass (2/3)mg and the short arm will be (1/3)mg. We do not have a simple method of figuringout which way the normal points. As with all pins, we consider it as two forces one vertical and onehorizontal.

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 theobject.

We will use Method A for the torques since that method is easiest to apply here. We will take the nail asthe pivot point since this eliminates the torques from the nail.

r F direction τz = rFsinθ

0 Nx 0

0 Ny 0L/2 (1/3)mg ½πθ CW mgLsin(½πθ)/6L (2/3)mg θ CCW 2mgLsinθ/3

Since Στz = 0, the equation we get is

mgLsin(½πθ)/6 + 2mgLsinθ/3 = 0.

Eliminating common terms and noting sin(½πθ) = cosθ, this becomes

cosθ/2 + 2sinθ = 0,

or

sinθ/cosθ = 1/4.

Using the identity, tanθ = sinθ/cosθ, we thus have θ = tan1(1/4) = 14.0°. The long side makes a 14.0°angle with the vertical.

9. A uniform 400 N boom is supported as shown in the figure below. Find the tension in the tie rope and theforce 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 theabsence 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 freebody diagram, an FBD where thelocation of the all forces are indicated so that torques can be calculated. Then we determine the threeequations 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 fromthe centre of mass, and the two tensions. We are given T2. The CM is obviously at the centre of theboom. We do not have a simple method of figuring out which way the normal points, instead we considerit 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 andangles are relatively easy to find. We will take the pin as the pivot point since this eliminates the torquesfrom the pin.

r F θ direction τz = rFsinθ

0 Px 0

0 Py 0L/2 W 40° CW LWsin(40°)/2

(3/4)L T1 50° CCW 3LT1sin(50°)/4

L T2 40° CW LT2sin(40°)


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 is

P = 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 = tan1(Py/Px) = 44.3° to the horizontal. Note that the pin force is not pointed solely along the length of theboom as one might expect.

Big Point To Remember: Pin forces are not always directed in the obvious direction.

10. In the figure below, a mass of 500 kg is held motionless in the air by a 120kg boom and a rope. Find thetension 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 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 havea simple method of figuring out which way the normal points, instead we consider it as two forces onevertical and one horizontal.

Fx = 0 ΣFy = 0

Px T1cos(π/2θ) = 0 Py mg T2 T1sinθ = 0

These tell us that Px = T1cosθ and Py = mg + T2 + T1sinθ. We are given the mass of the load so weknow T2 = mloadg = 4905 N.

We will use Method A for the torques since that method is easiest to apply here since the distances andangles are relatively easy to find. We will take the pin as the pivot point since this eliminates the torquesfrom the pin.

r F θ direction τz = rFsinθ

0 Px 0

0 Py 0L/2 mg π/2 φ CW Lmgsin(π/2φ)/2L T1 θ φ CCW LT1sin(θφ)

L T2 π/2 φ CW LT2sin(π/2φ)


Lmg[sin(π/2φ)]/2 + LT1sin(θφ) LT2sin(π/2φ) = 0 .

Eliminating L from the above, multiplying through by 2, and using the identity that sin(π/2φ) = cosφ yields,

mgcosφ + 2T1sin(θφ) 2T2cosφ = 0.

We rearrange to get T1 by itself,

T1 = (mg + 2T2)cosθ / 2sin(θφ) .

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 is

P = 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 =tan1(Py/Px) = 46.3° to the horizontal. Note that the pin force is not pointed solely along the length of theboom as one might expect.

11. A rectangular sign of mass 50.0 kg and width w = 5.00 m and l = length 4.00 m is hanging from a hingeand 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 forces that we know are working on the sign are a normal from the pin, the weight which acts fromthe centre of mass, and the tension. The CM is obviously at the centre of the sign. We do not have asimple method of figuring out which way the normal points, instead we consider it as two forces onevertical 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 theforces 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 yFx0 0 Px Py 0w 0 Tsinθ Tcosθ wTcosθw/2 l/2 0 mg wmg/2


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 Pxindicates we were wrong in assuming the the hinge pushed the sign to the right, it actually pulls the sign tothe left.

Thus the tension in the rope is 580 N and the horizontal and vertical components of the pin force are 526N and 245 N respectively.

12. Find the centre of mass of the object shown below. Determine the tension in the strings and the unknownangle 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. Wewill set the origin at the upper right corner where the string is attached. Note the symmetry of the object issuch that the CM must be located along a vertical line through the centre, that is the CM must be locatedin the y axis and that xcm = 0.16 m and zcm = 0.

Piece Mass yi (m) miyitop M/3 0.16 0.053333Mleft M/3 0.48 0.16Mright M/3 0.48 0.16MTotals: M 0.373333M

Thus ycm = (Σmiyi)/Mtotal = 0.373333 m.


The forces that we know are working on the sign are the two tensions and the weight which acts from thecentre of mass.

ΣFx = 0 ΣFy = 0

T1sinθ T2sin(65°) = 0 T1cosτq + T2cos(65°) Mg = 0

These 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 theforces easy to find. We will locate the pivot at the upper right corner because we have two unknownsthere.

x(m) y( m) Fx Fy τz = xFy yFx0 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


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, and

T1cosτ = 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 theunknown angle is θ = 16.4°. Substituting the angle back into either of the two equations yields the tensionin the right string T1 = 1.124 N.

13. 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. Thelength, l, is 12 cm.

To find the CM, we break the sign into two pieces each having all its mass located at their geometriccentres. Since the sign is uniform, its mass is proportional to its area. The total area of the sign is 9l2. Thecrosspiece has an area of 5l2 while the area of the vertical piece is 4l2. If the sign has mass M, thecrosspiece therefore has a mass of (5/9)M and the vertical piece a mass of (4/9)M. We will set the originat the hinge Note the symmetry of the object is such that the CM must be located along a vertical linethrough 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) miyitop (5/9)M ½l (5/18)Mlbottom (4/9)M 2l (8/9)MlTotals: M (11/18)Ml

Thus ycm = (Σmiyi)/Mtotal = (11/18)l.


The forces that we know are working on the sign are the tension, and the weight which acts from thecentre 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 theforces easy to find. We will locate the pivot at the hinge because we have two unknowns there.

x y Fx Fy τz = xFy yFx0 0 Hx Hy 05l l Tsin(40°) Tcos(40°) 5lTcos(40°) lTsin(40°)(5/2)l (11/18)l 0 Mg (5/2)lMg


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.

14. A ladder is propped against a wall making an angle with the floor. The wall is frictionless but thecoefficients of friction for the floor are μs and μk respectively. Obtain an expression for the smallest thatcan be if the ladder is not to slip. Recall that tanθ = sinθ/cosθ.


The forces that we know are working on the ladder are the weight which acts from the centre of mass, thenormal forces from the wall and floor, and friction. Since the ladder is not moving, we are dealing withstatic friction. Since we want the smallest angle, we are dealing with fs MAX. Since the ladder has atendency to move to the left, friction points to the left.

ΣFx = 0 ΣFy = 0

fs MAX Nw = 0 Nf mg = 0

These tell us that fs MAX = μNw and Nf = mg. We also know that fs MAX = μsNf where Nf is the normalforce 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 andangles 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/2θ CW ½Lmgsin(π/2θ)L Nw θ CCW LNwsinθ

Since τz = 0, the equation we get is

½Lmgsin(π/2θ) + LNwsinθ = 0 .

Eliminating L and noting that sin(π/2θ) = cosθ yields,

½mgcosθ + Nwsinθ = 0 .

The force equations gave Nw = μsmg, so we have

½mgcosθ + μsmgsinθ = 0 .

Rearranging and using tanθ = sinθ /cosτ, we get

θ = tan1(1 / 2μs) .

If the angle were any smaller than this, the ladder would slip.

15. A truss is made by hinging two 3.0m long uniform planks, each of weight 150 N, as shown below. Theyrest 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 arereflections of one another: to solve the problem, we need only consider one plank. However doing thismeans that we need to consider the force that one plank exerts on the other. It is a normal force, and byNewton's Third Law, must be equal but opposite on each. This requires the normal force to be horizontalas shown in the FBD of the left plank below. Also note that each plank supports half the load since theyare identical.

The problem mentions forces and looking at the diagram shows that the object would collapse in theabsence 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 freebody diagram, an FBD where thelocation of the all forces are indicated so that torques can be calculated. Then we determine the threeequations 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, thenormal forces from the wall and other plank, the load, and tension.

ΣFx = 0 ΣFy = 0

T Nplank = 0 Nf W ½Wload = 0

These tell us that T = Nplank and Nf = W + ½Wload = 400 N. A little trigonometry tells us that θ = cos1(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 andangles are easy to find. We will locate the pivot at the top of the plank because we have two unknownsthere.

r (m) F (N) θ direction τz = rFsinθ

0 Nplank 0

0 ½Wload 0

3 Nf π/2θ CW 3Nfsin(π/2θ)

2.5 T θ CCW 2.5Tsinθ1.5 W π/2+θ CCW 1.5mgsin(π/2+θ)


3Nfsin(π/2θ) + 2.5Tsinθ + 1.5Wsin(π/2+θ) = 0 .

We know that sin(π/2θ) = sin(π/2+θ) = cosθ and we already determined that Nf = W + ½Wload, so ourtorque 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.

16. A wheel of mass M and radius R rests on a horizontal surface against a step of height h (h < R). Ahorizontal 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. Thisindicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems bysketching the extended freebody diagram, an FBD where the location of the all forces are indicated sothat 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, thenormal forces from the wall and from the step, and the applied force. We do not know the direction of thenormal 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 andthus Ny = Mg.

We will use Method B for the torques since that method is easiest to apply here since the location of eachforce 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 thediagram. The x location is the same for each but takes a little work as shown in the diagram below whereit can be seen that x = [R2 (Rh)2]½.

x y Fx Fy τz = xFy yFx0 0 Nx Ny 0

[R2 (Rh)2]½ Rh 0 Mg [R2 (Rh)2]½Mg

[R2 (Rh)2]½ Rh F 0 (Rh)F

[R2 (Rh)2]½ h 0 Nf [R2 (Rh)2]½Nf


[R2 (Rh)2]½Mg (Rh)F [R2 (Rh)2]½Nf = 0 .

As pointed out, the wheel just loses contact with the ground when Nf = 0. That gives us our expressionfor F,

F = Mg [R2 (Rh)2]½ / (Rh) .

For any applied force less than this value, the wheel remains in contact with the ground.

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