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Heinemann Physics 1 4e

Chapter 11 Equilibrium of forces

Section 11.1 Torque

Worked example: Try yourself 11.1.1

CALCULATING TORQUE

A force of 255 N is required to apply a torque on a sports car steering wheel as it turns left. The force is applied at 90° to the 15.5 cm radius of the steering wheel. Calculate the torque on the steering wheel.

Thinking Working

Identify the variables involved and state them in their standard form.

T = ?

r⊥ = 0.155 m

F = 255 N

Apply the equation for torque. Rearrange if necessary. τ = r⊥F

T = 0.155 × 255

= 39.5 N m

State the answer with the appropriate direction. T = 39.5 N m anticlockwise


CALCULATING PERPENDICULAR DISTANCE

A truck driver can apply a maximum force of 1022 N on a large truck wheelnut spanner that has a length of 80.0 cm. The force is applied at 90° to the spanner. If the truck’s wheelnuts need a torque of 635 N m to make them secure, work out whether the spanner is long enough for the job.

Thinking Working


T = 635 N m

r⊥ = ?

F = 1022 N

Apply the equation for torque. Rearrange if necessary. τ = r⊥F

r⊥ = τF

= 635

1022

= 0.621 m

State the answer with the appropriate units. r⊥ = 62.1 cm

Compare the answer with the length of the spanner and state whether it is or isn’t appropriate for this task.

As the spanner is 80 cm, this is long enough to provide the minimum perpendicular distance of 62.1 cm. So the spanner is long enough.




CALCULATING TORQUE FROM THE PERPENDICULAR COMPONENT OF FORCE

A mechanic uses a 17.0 cm long spanner to tighten a nut on a winch. He applies a force of 104 N at an angle of 75.0° to the spanner.

Calculate the magnitude of the torque that the mechanic applies to the nut.

Give your answers to three significant figures.

104 N

75.0°

Thinking Working

Use the trigonometric relationship F⊥ = F sin θ to determine the force perpendicular to the spanner.

F⊥ = F sin θ= 104 sin 75.0°

= 100.5 N

Convert variables to their standard units. r = 17.0 cm

= 0.170 m

Apply the equation for torque.

τ = rF⊥τ = rF⊥

= 0.170 × 100.5

= 17.1

State the answer with the appropriate units. τ = 17.1 N m


CALCULATING TORQUE FROM THE PERPENDICULAR COMPONENT OF DISTANCE

A mechanic uses a 17.0 cm long spanner to tighten a nut on a winch. He applies a force of 104 N at an angle of 75.0° to the spanner.

Using the perpendicular distance, calculate the magnitude of the torque that the mechanic applies to the nut.


104 N

75.0°

Thinking Working

Convert variables to their standard units. r = 17.0 cm

= 0.170 m

Use the trigonometric relationship r⊥ = r sin θ to determine the perpendicular distance from the pivot point to the line of action of the force.

r⊥ = r sin θ= 0.170 sin 75.0°

= 0.164 m

Apply the equation for torque.

τ = r⊥Fτ = r⊥F

= 0.164 × 104

= 17.1

State the answer with the appropriate units. Note that this is the same answer as in the Worked example: Try Yourself ‘Calculating torque from the perpendicular component of force’.

τ = 17.1 N m



Section 11.1 ReviewKEY QUESTIONS SOLUTIONS

1 a The magnitude of the torque produced by a given force is proportional to the length of the force arm. By pushing the door at the handle, rather than the middle, the length to the force arm is increased.

b A crowbar can be used to generate a large torque because the force can be applied at a large distance from the pivot.

2 a τ = r⊥F

= 2 × 100

= 200 N m anticlockwiseb The torque is zero because the line of action passes through the pivot point.

3 r⊥= τF

= 15 ÷ 30

= 0.5 m

4 F = τr⊥

= 9

0.50

= 18 N

5 τ = r⊥F

= 0.40 × 225

= 90 N m

6 a T = r × F = 0.5 × 1.0 × 9.8 = 4.9 N m b T = r × F = 1.0 × 1.0 × 9.8 = 9.8 N m c T = r × F sin θ = 1.0 × 1.0 × 9.8 × sin 30° = 4.9 N m

7 T = rF sin θ= 0.30 × 300 × sin 30

= 0.30 × 300 × 0.5

= 45 N m

8 a Weight of skip: Fg = mg = 3500 × 9.80 = 3.4 × 104 Nb The effective force arm remains at 15 m throughout, so the torque does not change.c τ = r × F sin θ = 25 × 3.43 × 104 × sin 37° = 5.2 × 105 N m clockwise about the pivot.

(Or simply find: τ = r × F = 15 × 3.43 × 104 = 5.2 × 105 N m clockwise about the pivot.)



11.2 Translational equilibrium


CALCULATING TRANSLATIONAL EQUILIBRIUM IN ONE DIMENSION

Three cars are parked on a beam bridge that has a mass of 500 kg. The left pillar (labelled X) applies a force of 2.00 × 104 N upwards. If the situation is in translational equilibrium then calculate the force provided by the right-hand pillar (labelled Y). Use g = 9.80 N kg−1 when answering this question.

1000 kg

X Y

1500 kg 2000 kg

Thinking Working

Identify the variables involved and state them with their directions in their standard form.

m1 = 1000 kg

m2 = 1500 kg

m3 = 2000 kg

mB = 500 kg

FX = 20 000 N up

g = 9.80 N kg−1 down

Apply a sign convention to the vector data. FX = +20 000 N

g = −9.80 N kg−1

Identify the object that is in translational equilibrium. This is the object on which all the forces are acting.

The object experiencing translational equilibrium is the bridge.

Apply the equation for translational equilibrium in one dimension.

F(net,y) = 0

Expand the equation to include each of the forces acting on the beam.

F1 + F2 + F3 + FB + FX + FY = 0

m1g + m2g + m3g + mBg + FX + FY = 0

Substitute the data into the equation and solve for the unknown.

1000 × −9.80 + 1500 × −9.80 + 2000 × −9.80 + 500 × −9.80 + 20 000 + FY = 0

−9800 + −14 700 + −19 600 + −4900 + +20 000 + FY = 0

−29 000 + FY = 0

FY = 29 000 N

State the answer with the appropriate direction. FY = 29 000 N up




CALCULATING TRANSLATIONAL EQUILIBRIUM IN TWO DIMENSIONS

A concrete beam of mass 1500 kg is being lifted by cables labelled 1 and 2, as shown in the diagram. The beam is moving upwards with a constant velocity of 2.0 m s−1. Calculate the tension in cable 1 and cable 2. Ignore the mass of the cable and use g = 9.80 N kg−1 when answering this question.


cable 1 cable 2

60°

60° 60°

1500 kg

Thinking Working

Construct a vector diagram adding all of the forces together. Fnet= 0

60°

60°

FT2

FT1

FgB

Apply horizontal components and vertical components.

60°

60°

FT2 F2V

F2H

F1H

F1VFT1

FgB

Fnet= 0

Recognise that, in the horizontal dimension, F2H is in equilibrium with F1H.

F2H

F1H

Fnet= 0



Recognise that, in the vertical dimension, FgB is in equilibrium with FRV and FLV.

F2V

F1V

Fnet= 0

FgB

Apply the equation for translational equilibrium in the vertical dimension. Recognise that F1V and F2V are equal in magnitude and therefore each is half of FgB.

F(net,V) = 0

F1V = F2V

Expand the equation to include each of the vertical forces acting on the sign.

FgB + F1V + F2V = 0

mBg + F1V + F2V = 0


1500 × −9.80 + F1V + F2V = 0

−14 700 + F1V + F2V = 0

F1V + F2V = 14 700

F1V = F2V = 7350 N

Draw the right triangle with one of the vertical components of the tension and the angle. 60°

FT2 F2V

Use trigonometry to solve for the tension in one of the cables, which will equal the tension in the other cable as well.

FT2 = 7350

sin 60°

= 8487 NFT2 = FT1 = 8487 N


1 D. As the net force is zero the translational acceleration must also be zero. A and B are incorrect, as both conditions (stationary and moving at constant velocity) are possible.

2 For a body to be in translational equilibrium, the vector sum of the forces acting on it must be zero. This situation applies to any stationary body or to one moving with constant velocity. Options A, B and D are therefore examples of translational equilibrium.

3 a 150 N up

50 N

150 N up200 N

b 40 N west

65 N

40 N west25 N

c zero

10 N

ΣF = 010 N

d ∑F= √(1022 + 102) = 14 N north west

10 N

10 N

ΣF



4 Fnet = 0

Fg + FT = 0

mBFg + FT = 0

0.355 × −9.80 + FT = 0

−3.479 + FT = 0

FT = 3.479 N up

5 Fnet = 0

FgT + FFJ + FgP + 4FT = 0

mTg + mJg + mPg + 4FT = 0

79 × −9.80 + 68 × −9.80 + 225 × −9.80 + 4FT = 0

−774.2 + −666.4 + −2205 + 4FT = 0

−3645.6 + 4FT = 0

4FT = 3645.6

FT = 3645.6

4

FT = 911 N

6 a In the x-direction:

F(net,x) = 400 cos 40° + 400 cos 40°

F(net,x) = 613 N

In the y-direction:

F(net,y) = 400 sin 40° + −400 sin 40°

F(net,y) = 0

So the net force is 613 N to the right.b Translational equilibrium equation:

Fnet = 0

613 + Ffr = 0

Ffr = −613 N

So the friction force is 613 N to the left.

7 Downwards force on sign = 5.0 × 9.80 = 49 N

Each vertical component of the tension forces in the wires is equal to 24.5 N to balance the downward force on the sign due to gravity.

cos 40° = 24.5

Ft

Ft = 24.5

cos 40° = 32 N



Section 11.3 Static equilibrium


CALCULATING STATIC EQUILIBRIUM

A set of scales (with one longer arm) is used to measure the mass of gold. A lump of gold with a mass of 150 g is placed on the short arm, which is 10.0 cm long, and a standard set of masses are placed on the long arm.

Use g = 9.80 N kg−1 when answering these questions. Give your answers to three significant figures.

150grams

150grams

a Calculate the force applied to the scale’s arm due to the pivot point if a standard mass of 50.0 g exactly balances the gold.

Thinking Working


mg = 0.150 kg

mm = 0.050 kg


Apply a sign convention to the vector data g = −9.80 N kg−1

Identify the object that is in translational equilibrium. This is the object on which all the forces are acting.

The object experiencing translational equilibrium is the scale arm.

Apply the equation for translational equilibrium in the vertical dimension.

F(net,y) = 0

Expand the equation to include each of the forces acting on the scales arm.

Fg + Fm + FP = 0mgg + mmg + FP = 0


(0.150 × −9.80) + (0.050 × −9.80) + FP = 0

(−1.47) + (−0.49) + FP = 0

−1.96 + FP = 0

FP = 1.96 N

State the answer with the appropriate direction. FP = 1.96 N up



b Calculate how long the long arm should be in order to balance the gold.

Thinking Working


mg = 0.150 kg

mm = 0.050 kg

r⊥g = 0.10 m


Identify the object that is in rotational equilibrium. This is the object on which all the torques are acting.

The object experiencing rotational equilibrium is the scale arm.

Decide the reference point about which the torques will be calculated.

The reference point is the pivot.

Decide which force causes the clockwise torque and which force causes the anticlockwise torque around the chosen reference point.

The force of the standard mass on the scale arm provides the clockwise torque.

The force of the gold on the scale arm provides the anticlockwise torque.

Apply the equation for rotational equilibrium. ∑τclockwise = ∑τanticlockwise

Expand the equation to include each of the torques acting on the balance.

r⊥mFm = r⊥gFg


Fm r⊥m = Fg r⊥g

mm gr⊥m = mg gr⊥g

0.050 × 9.80 × r⊥m = 0.150 × 9.80 × 0.10

r⊥m = 0.150 × 9.80 × 0.10

0.050 × 9.80

= 0.30 m




CALCULATING STATIC EQUILIBRIUM USING A DIFFERENT REFERENCE POINT

Verify that the see-saw plank in the figure below is also in rotational equilibrium about the reference point where the boy is sitting, at point Y. The weight of the boy and girl are 196 N and 294 N, respectively, and the force of the pivot on the plank is 490 N upwards. Assume that the plank’s mass is negligible.

1.5 m1.0 m

pivot point

FP = 490 N

Fboy = 196 N Fgirl = 294 N

Y X

Thinking Working


The reference point is the position of the boy.


FP = 490 N

Fg = 294 N

Fb = 196 N

r⊥g = 2.50 m

r⊥P = 1.50 m


The object experiencing rotational equilibrium is the see-saw.

Decide which forces cause the clockwise torques and which force causes the anticlockwise torques around the chosen reference point.

The force of the girl on the see-saw provides the clockwise torque.

The force of the pivot on the see-saw provides the anticlockwise torque.


Expand the equation to include each of the torques acting on the see-saw

Fg r⊥g = FP r⊥P


Fg r⊥g = FP r⊥P

294 × 2.50 = 490 × 1.50

735 = 735

Identify the magnitude of the clockwise torque compared to the magnitude of the anticlockwise torque.

Around reference point Y (the position of the boy), the clockwise torque due to the girl on the plank is equal to the anticlockwise torque due to the pivot point on the plank. So the plank is in rotational equilibrium.




CALCULATING STATIC EQUILIBRIUM WITH TWO UNKNOWNS

For the painter on the plank scenario above, determine the tension on the right-hand rope (Ft2).

Thinking Working


Note that this must be the point at which the other unknown force acts.

The reference point is the point at which the rope providing the tension force Ft1 is attached to the plank.


mpl = 20.0 kg

mpa = 70.0 kg

r⊥Ft2 = 6.00 m

r⊥c = 3.00 m

r⊥pa = 4.00 m

g = 9.80 N kg−1


The object experiencing rotational equilibrium is the plank.

Decide which forces cause the clockwise torques and which force causes the anticlockwise torque around the chosen reference point.

The force of the painter on the plank provides a clockwise torque.

The force of gravity on the plank provides another clockwise torque.

The tension force of the right hand rope on the plank provides the anticlockwise torque.


Expand the equation to include each of the torques acting on the plank.

r⊥Ft2 Ft2 = r⊥c Fpl + r⊥pa Fpa


Ft2 × 6.00 = 20.0 × 9.80 × 3.00 + 70.0 × 9.80 × 4.00

Ft2 = 20.0 × 9.80 × 3.00 + 70.0 × 9.80 × 4.00

6.00= 588 + 274

6.00

= 555 N




USING STATIC EQUILIBRIUM TO CALCULATE THE FORCE ON A CANTILEVER

Determine the magnitude and direction of the force that the left-hand support must supply so that the beam is in static equilibrium (F1).

8.0 m 10.0 m

r2 = 8.0 m

Fb = 2940 N

F2

F1

rcm = 9.0 m

(a) (b)

mb = 300.0 kg

Thinking Working



The reference point is the point at which the right-hand support is attached to the beam.


mb = 30.0 kg

r⊥F1 = 8.00 m

r⊥c = 1.00 m

g = 9.80 N kg−1


The object experiencing rotational equilibrium is the beam.


The force of gravity on the beam provides the clockwise torque.

The force of the left-hand support on the beam provides the anticlockwise torque.


Expand the equation to include each of the torques acting on the plank.

Fbr⊥c = F1r⊥F1


30.0 × 9.80 × 1.00 = F1 × 8.00

F1 = 30.0 × 9.80 × 1.00

8.00

= 36.8 N

State the direction of the force acting on the object in equilibrium.

The force is downwards on the beam.




USING STATIC EQUILIBRIUM TO CALCULATE THE TENSION IN A TIE THAT IS SUPPORTING A BEAM

A uniform 5.00 kg beam, 1.80 m long, extends from the side of a building and is supported by a wire tie which is attached to the beam 1.20 m from a hinge (h) at an angle of 45°. Calculate the tension (Ft) in the wire that is supporting the beam.


1.8 m

1.2 m

45°

h

Thinking Working



The reference point is the point at which the hinge (h) is attached to the beam.


Here the subscripts used are: b = beam, c = centre of mass of the beam and w = wire

mb = 5.00 kg

r⊥c = 0.90 m

rw = 1.20 m

g = 9.80 N kg−1


The object experiencing rotational equilibrium is the beam.


The force of gravity on the beam provides the clockwise torque.

The force of tension in the wire tie on the beam provides the anticlockwise torque.


Expand the equation to include each of the torques acting on the beam.

Fb r⊥c = Ft r⊥w

Solve for the perpendicular distances from the force arm to the line of action of the force.

r⊥w = rw sin 45°

= 1.20 × sin 45°

= 0.849 m

Substitute the data into the equation and solve for the unknown force.

5.00 × 9.80 × 0.90 = Ft × 1.20 × sin 45°

Ft = 5.00 × 9.80 × 0.90

1.20 × sin 45°

= 52.0 N


1 The bench will not work successfully. It is not balanced and will topple over because the weight vector from the centre of gravity is outside the base provided by the two supports. To improve the design the student needs to move the supports apart so the centre of gravity is between the two supports.

2 D. As the net torque is zero then rotational equilibrium must be established by having equal clockwise and anticlockwise torques.

3 C. The front cog is not rotating and the cyclist is travelling at constant velocity. A, B and D are incorrect as the wheels must be turning in order for the bike to be moving.

4 r⊥a Fc = r⊥a Fc

r⊥a 75.0 × 9.80 = 2.25 × 25.0 × 9.80

r⊥a = 2.25 × 25.0 × 9.80

75.0 × 9.80

r⊥a = 0.750 m



5 Let the right-hand side support be the reference point, then:

r⊥L FL = r⊥100 F100 + r⊥150 F150 + r⊥200 F200

r⊥L−R FL = r⊥100−R F100 + r⊥150−R F150 + r⊥200−R F200 + r⊥shelf Fshelf

10.0 × FL = (7.00 × 100 × 9.80) + (3.00 × 150 × 9.80) + (1.50 × 200 × 9.80) + (5.0 × 50.0 × 9.80)

FL = 6860 + 4410 + 2940 + 2450

10.0= 1666 N

Let the left-hand support be the reference point, then:r⊥R FR = r⊥100 F100 + r⊥150 F150 + r⊥200 F200 + r⊥shelf Fshelf

FR × 10.0 = (3.0 × 100 × 9.80) + (7.0 × 150 × 9.80) + (8.5 × 200 × 9.80) + (5.0 × 50.0 × 9.8)

FR = 3234 N

6 a FX acts downwards so that torques balance.b Let X be the reference point. The weight force acting at the awning’s centre of gravity provides the clockwise torque,

and the force of the support at Y provides the anticlockwise torque.

∑τclockwise = ∑τanticlockwise

r⊥c Fc = r⊥Y FY

2.0 × 900 × 9.80 = 1.8 × FY

FY = 9800 Nc Let Y be the reference point. The weight force acting at the awning’s centre of gravity provides the clockwise torque,

and the force of the support at X provides the anticlockwise torque.

∑τclockwise = ∑τanticlockwise

r⊥c Fc = r⊥X FX

0.2 × 900 × 9.80 = 1.8 × FX

FX = 980 N

7 r⊥aFa = r⊥c1Fc1+ r⊥c2Fc2

r⊥a = (1.50 × 20.0 × 9.80) + (2.50 × 20.0 × 9.80)

90.0 × 9.80

= 294 + 490

882

= 0.889 m

8 a Fh = 800 cos 30° = 400 N

Fv = 800 sin 60° = 693 Nb Taking torques around the base of the post:

(Ft × 1.0) − (400 × 0.85) = 0

Ft = 340 N

Chapter 11 Review1 B. As the force or the force arm (lever arm length) increases, or both increase, then the torque will increase.

2 τ = r⊥F= 0.200 × 25.0

= 5.00 N m

3 r⊥ = τF

= 3.47

12.0

= 0.289 m

4 D. The combination of force and the force arm length provides a greater torque than the other options.

5 τ = r⊥F= 3.00 × 5000

= 15 000 N m

6 r⊥ = τF

= 32.1

24.0

= 1.34 m



7 C. The maximum effect is achieved if the force applied is perpendicular (at 90°) to the surface of the object.

8 τ = rF⊥= rF sin θ= 2.00 × 30 sin 40.0°

= 38.6 N m

9 τ = rF⊥= rF sin θ= 0.450 × 62.0 sin 65.0°

25.3 N m

10 A. Either of the perpendicular components can be used and not only one or the other. Both aren’t necessary.

11 a r⊥ = r sin θ= 1.50 × sin 30.0°

= 0.750 mb τ = r⊥F

T = 0.750 × 12.5

= 9.375 N m

12 τ = r⊥F

= 0.0700 × 8.50

= 0.595 N m

13 τ = rF sin θ= 0.900 × 82.0 sin 50.0°

= 56.5 N m

14 Consider the left-hand side, then:

Fv = FT sin θ= 40 sin 40°

= 25.7 N

The vertical component of the tension on the right hand side is also 25.7 N.

Total forces upwards (vertically) = 25.7 + 25.7 = 51.4 N.

The mass of the picture cannot exceed:

Fg = mg

51.4 = m × 9.8

m = 5.25 kg

15 Fnet,Y = 0

FgP + FLV + FRV = 0

75.0 × −9.80 + FLV + FRV = 0

−735 + FLV + FRV = 0

FLV = FRV = 735

2= FRV = 367.5 N

sin θ= FRV

FTR

FTR = FRV

sin θ

= 367.5

sin 10°

= 2116 N

FTR = FTL = 2116 N

16 Fnet,Y = 0

Fg = FT

mg = 7.50

m = 7.50

9.80

= 0.765 kg



17 Fnet,Y = 0

FgBB + FTAV = 0

mBBg + FTAV = 0

100 × −9.80 + FTAV = 0

FTAV = 980 N

sin 60° = FTAV

FTA

FTA = 980

sin 60°

= 1132 N∑Fx = 0

FTB + FTAH = 0

FTB + FTA cos 60° = 0

FTB + 1132 × cos 60° = 0

FTB = −566 N (566 N opposing A)

18 C. Because there is no net torque about the reference point and therefore rotation does not occur when the object is in rotational equilibrium.

19 B. For an object to experience static equilibrium it must experience both rotational equilibrium and translational equilibrium.

20 A. As Tom’s velocity is not changing the translational acceleration must also be zero.

21 a In this solution, we have used a negative sign to denote anticlockwise torques.

F1 = m1g

25 × 9.80 = 245 N

τ1 = r1F1

= −0.6 × 245

= −147 N m

= 147 N m anticlockwiseb F2 = m2g

= 20 × 9.80

= 196 N

τ2 = r2F2

= 1.0 × 196

= 196 N m

= 196 N m clockwisec τnet = τ1 + τ2

= −147 + 196

= 49 N m

= 49 N m clockwise

22 τnet = τ1 + τ2

= r1W1 + r2W2

= −1.5 × 50 + 1.5 × 100

= −75 + 150

= +75 N m

= 75 N m clockwise

23 τ = r⊥F

= 0.120 × 30

= 3.60 N m

24 τ = rF sin θ= 1.35 × 64.0 sin 60.0°

= 74.8 N m



25 τ = rF sin θ= 1.05 × 50.0 × sin 78.0°

= 51.4 N m

26 Fnet = 0

FW + FT = 0

mc g + FT = 0

(86.5 × −9.80) + FT = 0

−847.7 + FT = 0

FT = 847.7 N

27 Fnet = 0

Fgt + Fgf + 4FL = 0

mtg + mfg + 4FL = 0

40.0 × −9.80 + 4.50 × −9.80 + 4FL = 0

−392 + −44.1 + 4FL = 0

−436.1 + 4FL = 0

4FL = 436.1

FL = 436.1

4

= 109 N

28 Fnet,Y = 0

FLy + FRy + msg = 0

FL sin 35° + FR cos 70° + 12.5 × −9.80 = 0

0.5736FL + 0.3420FR + −122.5 = 0

0.5736FL + 0.3420FR = 122.5

Fnet,X = 0

−FLx + FRx = 0

−FL cos 35° + FR sin 70° = 0

−0.8191FL + 0.9397FR = 0

0.9397FR = 0.8191FL

FR = 0.8191FL

0.9397

= 0.8717FL

0.5736FL + 0.3420FR = 122.5

0.5736FL + 0.3420 × 0.8717FL = 122.5

0.5736FL + 0.2981FL = 122.5

0.8717FL = 122.5

FL = 122.5

0.8717

= 140.5 N

FR = 0.8717FL

= 0.8717 × 140.5

= 122.5 N



29 Fnet,Y = 0

msg + mbg + meg + FT = 0

(0.145 × −9.80 + 0.0100 × −9.80 + 0.0225 × −9.80) + FT = 0

−1.421 + −0.0980 + −0.2205 + FT = 0

FT = 1.7395 N

With the rod in static equilibrium and the Sun as the reference point:∑Tcw = ∑Tacw

r⊥rFr + r⊥EFE = r⊥TFT

(0.500 × 0.01 × 9.8 ) + (1.00 × 0.0225 × 9.8 ) = (rT × 1.7395)

(0.500 × 0.0980) + (1.00 × 0.2205) = (r⊥T × 1.7395)

0.0490 + 0.2205 = (r⊥T × 1.7395)

r⊥T = 0.2695

1.7395

= 0.1549 m

= 15.5 cm from the Sun

30 a i The three forces acting on the left-hand cantilever are the weight of the beam (3920 N), the force from pillar A (FA) acting downwards and the force from pillar B (FB) acting upwards: FA + 3920 = FB

Taking torques around the left-hand end of the left cantilever:(FB × 1.2) − (3920 × 1.5) = 0

FB = 5880

1.2 = 4.9 kN up

So: FA = 980 N downBy symmetry: FC = 4.9 kN up and FD = 980 N down

ii The force that the pedestrian FP exerts on the end of the cantilever is 35 × 9.8 = 343 N down.

FA + 3920 + 343 = FB

Taking torques around the left-hand end of the left cantilever:(FB × 1.2) − (3920 × 1.5) − (343 × 3.0) = 0

FB = 6909

1.2 = 5758 = 5.8 kN up

So: FA = 1495 N = 1.5 kN downBy symmetry: FC = 5.8 kN up and FD = 1.5 kN down

b As the woman walks from A to B, the force acting in pillar A decreases and the force acting in B increases. When the woman passes point B and continues on to point P, the forces in both A and B increase in order to produce a greater torque to counterbalance the increase in torque as she moves to point P.

heinemann pic 1 4e chapter 11 equilibrium of forces · heinemann pic 1 4e chapter 11 equilibrium of...

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