PHY–302 K. Solutions for Problem set # 10. Non-textbook...

PHY–302 K. Solutions for Problem set # 10.

Non-textbook problem #1

FB

FCmg

LB LC

L

The equilibrium conditions for any rigid body are

∑

Fx = 0,∑

Fy = 0,∑

τ = 0. (1)

The stretcher in question is subject to three forces: The patient’s weight Mg, force FB of

Bob’s hands, and force FC from Charlie’s hands. All these forces are vertical, so∑

Fx = 0 is

trivially true, and there is only one non-trivial balance-of-forces equation,

∑

Fy = FB + FC − Mg = 0. (2)

In the balance-of-torques equation∑

τ = 0, we may treat any point P we like as a pivot,

as long as we calculate toques of all forces with respect to the same pivot point P. For the

problem at hand, it’s convenient to put P at one end of the stretcher, for example at the front

end where Bob holds the stretcher. For this choice of a pivot, the force FB has zero lever

arm and hence zero torque. On the other hand, the force FC acts at the other end of the

stretcher, so its lever arm is L = 8 ft — the full length of the stretcher — and the torque is

τ(FC) = −FC ×L, where the ‘−’ sign indicates the counterclockwise direction of this torque.

Finally, the patient’s weight Mg is distributed all over the patient’s body, but for the

purpose of calculating the torque we may treat it as acting at the patient’s center of gravity

(same as his center of mass). This center of mass lies at LB = 3 ft from the pivot point (Bob’s

hands), so the lever arm of Mg is LB and the torque is τ(Mg) = +Mg × LB, where the ‘+’

sign indicates the clockwise direction of this torque.

1

Altogether, the net torque around out chosen pivot point is

∑

τ = τ(FB) + τ(FC) + τ(Mg) = 0 + FC × L − Mg × LB . (3)

Demanding that this net torque cancels out, we have

FC × L − Mg × LB = 0 (4)

and therefore

FC = Mg ×LB

L= 200 lb×

3 ft

8 ft= 80 lb. (5)

Consequently, according to the balance-of-forces equation (2),

FB = Mg − FC = Mg − Mg×LB

L= Mg×

(

1 −LB

L=

LC

L

)

= 200 lb×5 ft

8 ft= 120 lb.

(6)

Thus, Charlie carries 80 pounds of patient’s weight and Bob carries the remaining 120 pounds.

Textbook problem 9.12:

Since the traffic light is free to swing on the lower wire, the balance of torques is automatic,

and all we need to check is the balance of forces. The tension of the lower wire must be equal

to the light’s weigh Mg, while the tensions of the two upper wires follow from zero net force

on the junction of the three wires:

∑

~F = ~T1 + ~T2 + M~g = ~0, (7)

or in components,

0 =∑

Fx = −T1 cos θ1 + T2 cos θ2 ,

0 =∑

Fy = +T1 sin θ1 + T2 sin θ2 − Mg.(8)

To solve this equation system, let’s multiply the first equation by sin θ1, the second equation

2

by cos θ1, and then add the two equations. This gives us

−T1 cos θ1 × sin θ1 + T2 cos θ2×sin θ1 + T1 cos θ1 × sin θ1 + T2 sin θ2×cos θ1 −Mg×cos θ1 = 0

(9)

where terms involving the T1 cancel out. Consequently, the tension in the second (upper right)

wire is

T2 =Mg cos θ1

cos θ2 sin θ1 + sin θ2 cos θ1. (10)

The denominator here can be simplified using a trigonometric identity

sin(θ1 + θ2) = cos θ2 × sin θ1 + sin θ2 × cos θ1 , (11)

hence

T2 =Mg cos θ1sin(θ1 + θ2)

. (12)

As to the first wire, the first equation (8) gives us

T1 = T2 ×cos θ2cos θ1

=Mg cos θ2sin(θ1 + θ2)

. (13)

For the traffic light in question, Mg = 323 N ≈ 73 lb, θ1 = 53◦, θ2 = 37◦, θ1 + θ2 = 90◦,

hence

T1 = 323 N×cos 37◦

sin 90◦= 258 N ≈ 58 lb, T2 = 323 N×

cos 53◦

sin 90◦= 195 N ≈ 44 lb. (14)

3


There are 4 forces acting on the diving board, all vertical: the forces FA and FB of the two

supports, the diver’s weight Mg, and the platform’s own weight mg. Here is the force diagram:

A B C D

FA FB

Mg

mg

(15)

Note that the force at point A is directed down rather than up, so the board must be cemented

/ welded / glued / nailed to the support A rather than simply laying on top of it. Indeed, the

gravity forcesMg andmg create clockwise torques on the board with respect to the support B;

to balance these torques we need a counterclockwise τ(FA), and that’s why the FA must be

directed down rather than up.

Specifically, let’s use point B as a pivot and also as the origin of the x coordinate. Thus,

XB = 0, hence XA = −1.0 m, XD = +3.0 m, and

XC ≡ Xcm =XA +XD

2= +1.0 m. (16)

Consequently

∑

τ = FA ×XA + FB ×XB + mg ×XC + Mg ×XD (17)

where the first term is negative (counterclockwise) because xA < 0. Solving this equation for

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the FA, we find

FA =mg ×XC + Mg ×XD

−XA= +1.0×mg + 3.0×Mg. (18)

Numerically, FA ≈ 1700 N ≈ 380 lb if we don’t take the board’s own weight into account

(part (a)) and FA ≈ 2050 N ≈ 460 lb if we do (part (b)).

To find the force at the support B, we use the balance-of-forces equation. Since all forces

are vertical, there is only one equation∑

Fy = 0, and given the directions of all the forces, it

reads

−FA + FB − mg − Mg = 0 =⇒ FB = FA + mg + Mg. (19)

Combining this formula with eq. (18) for the FA, we have

FB = 2.0×mg + 4.0×Mg. (20)

Numerically, FA ≈ 2270 N ≈ 510 lb if we don’t take the board’s own weight into account

(part (a)) and FA ≈ 2960 N ≈ 670 lb if we do (part (b)).


Each of the three crossbars in figure 9–71 acts as balance scales. Here is the force diagram:

MLg MRg

T

LL LR

(21)

The net torque about the point where the upper wire is attached to the crossbar is

τnet = −LL ×MLg + 0× T + LR ×MRg. (22)

For a balanced crossbar, this net torque must be zero, hence the masses hanging from the

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crossbar must satisfy

LL ×ML = LR ×MR . (23)

Note that here ML and MR are net masses of all objects hanging from the appropriate end of

the crossbar. Thus, for the top crossbar ML = MA while MR = MB +MC +MD. Likewise,

for the middle crossbar MR = MB while ML = MC +MD. Finally, for the bottom crossbar

ML = MD and MR = MC . Consequently, eqs. (23) for the three crossbars read

30.00×MA = 7.50× (MB +MC +MD), (top)

15.00× (MC +MD) = 5.00×MB , (mid)

17.50×MD = 5.00×MC . (bot)

The problem gives us MB = 885 g. Using this value and eq. (mid), we find the net mass

of the two lower objects is MC +MD = 295 g. Combining this datum with eq. (bot), we have

17.50×MD = 5.00× (MC = 0.295 kg − MD) (24)

and hence

MD =5.00× 295 g

17.50 + 5.00= 65.6 g (25)

while MC = 295 g − 65.6 g = 229.4 g ≈ 229 g.

Finally, on the right hand side of eq. (top), MB +MC +MD = 885 g + 295 g = 1180 g.

Consequently,

MA =7.50× 1180 g

30.00= 295 g. (26)

6


(a) The figure 9–52 shows the beam is hinged to the wall. This allows the wall’s force Fw on

the beam to have any direction it needs to have, but it does not create any torque on the beam

(with respect to the hinge). Hence, the net torque on the beam is due to the wire tension T

(which has lever arm L× sin 40◦) and the beam’s weight mg (which has lever arm 12L), thus

τnet = 0 − T × L sin 40◦ + mg × 12L. (27)

Rotational balance of the beam requires this net torque to vanish, and therefore

T =mg × 1

2L

L sin 40◦=

mg

2 sin 40◦= 206 N ≈ 46 lb. (28)

(b) To calculate the force Fw at the hinge, we use the balance-of-forces equations

0 =∑

Fx = Fwx − T cos 40◦,

0 =∑

Fy = Fwy + T sin 40◦ − mg.(29)

Solving these equations, we find

Fwx = T cos 40◦ = 158 N ≈ 35 lb (30)

and

fwy = mg − T sin 40◦ = mg −mg

2=

mg

2= 132 N ≈ 30 lb. (31)

Converting the components of the Fw vector into its magnitude and direction, we have

|Fw| =√

F 2wx + F 2

wy = 206 N ≈ 46 lb (32)

and

θ(Fw) = arctanFwy

Fwx= 40◦. (33)

7


Let’s start by deriving the stability condition for a generic ladder of weight mg, length L, and

center of mass at a distance ν×L from the bottom. (For a symmetric ladder ν = 12.) Suppose

the ladder leans against a smooth (approximately frictionless) wall making angle θ with the

floor, and a painter of weight Mg is standing on the step at distance D from the bottom.

Here is the diagram of the ladder and all the forces acting on it:

CM

D

θmgMg

N

N ′

f

At the bottom of the ladder, there is a vertical normal force N from the floor and a horizontal

static friction force f . At the top, there is a horizontal normal force N ′ from the wall. In

principle, there should also be a vertical friction force f ′, but a smooth wall has a very small

friction coefficient, so f ′ is much smaller than the other forces and we are going to neglect it.

Finally, there are two vertical gravity forces: the weight Mg of the painter, and the weight

mg of the ladder itself.

As long as the ladder stays motionless, it has to satisfy three equilibrium conditions: the

balance of forces in both vertical and horizontal directions, and the balance of torques,

∑

Fx = N ′ − f = 0, (34)∑

Fy = N − Mg − mg = 0, (35)∑

τ = τ(N) + τ(f) + τ(N ′) + τ(Mg) + τ(mg) = 0. (36)

8

In the balance-of-torques equation (36), we may evaluate the torques relative to any pivot

point we like, as long as we use the same pivot point for all the forces. For convenience, one

should let the pivot point to lie on lines of as many unknown forces as possible, because this

will eliminate those forces from the torque equation. For the problem at hand, we put the

pivot at the bottom of the ladder. The forces N and f act at this point, so they have zero

lever arms and consequently

τ(N) = τ(f) = 0. (37)

As to the remaining forces, the normal force N ′ from the wall acts at distance L – the ladder’s

length — from our pivot point, and the direction of this force makes angle θ with the ladder

and 180◦ − θ with the radius-vector from the pivot. Consequently, the N ′ has lever arm

L× sin(180◦ − θ) = L× sin θ and the torque

τ(N ′) = +N ′ × L sin θ. (38)

The painter’s weight Mg acts at the step where he stands, which is as distance D from the

ladder’s bottom. The weight acts at angle 90◦ − θ from the ladder or 90◦ + θ from the

radius-vector. Consequently, it has lever arm D × sin(90◦ + θ) = D × cos θ and torque

τ(Mg) = −Mg ×D cos θ. (39)

Finally, the ladder’s own weight mg is distributed all along the ladder, but for the purpose of

calculating torques, we can take it as acting at the ladder’s center of mass. For a symmetric

ladder (no difference between the top and bottom ends), the center of mass lies in the middle,

at the distance 12L from the bottom. For an asymmetric ladder this distance would be different,

so to keep our analysis general let’s denote it ν × L. Again, the direction of mg makes angle

90◦ + θ with the radius-vector, so the lever arm is νL× cos θ and the torque is

τ(mg) = −mg × νL cos θ. (40)

Combining the torques (37) through (40) into the balance-of-torques equation (36), we arrive

9

at∑

τ = 0 + 0 + N ′ × L sin θ − Mg ×D cos θ − mg × νL cos θ = 0. (41)

Solving this equation for the normal force N ′ from the wall gives us

N ′ =cos θ

sin θ×

(

D

L×Mg + ν ×mg

)

. (42)

The remaining unknown forces — the normal force N from the floor and the friction force

f there — follow from the balance-of-forces equations (34) and (35):

f = N ′ =1

tan θ×

(

D

L×Mg + ν ×mg

)

, (43)

N = Mg + mg. (44)

The equations (42), (43), and (44) must hold true as long as the ladder does not move.

In addition, the friction force f and the normal force N at the floor must satisfy the static

friction rule

|f | ≤ µs ×N. (45)

For the forces given by eqs. (43) and (44), this inequality becomes

1

tan θ×

(

D

L×Mg + ν ×mg

)

≤ µs × (Mg +mg) (46)

or equivalently

µs × (M +m)× tan θ ≥D

L×M + ν ×m. (47)

In terms of the position D of the climber, this inequality amounts to

D ≤ Dmax = L×

(

µs × tan θ ×M +m

M− ν ×

m

M

)

. (48)

If the painter climbs any higher than this, the friction force (43) would violate the static

friction rule (45) for the normal force (44). This cannot happen, so the friction would not be

10

able to balance the force N ′ from the wall and / or the torque of the N ′ would not balance the

torque of the gravity forces. Consequently, the ladder would slide to the right and tilt down,

until it falls flat on the floor, and the painter falls on his face (or perhaps a less sensitive part

of his body).

Now let’s apply our generic stability conditions to the ladder in question. The problem

tells us M = 55.0 kg, m = 12.0 kg, and Dmax/L = 70% ≡ 0.70. Also, the figure 9–61 below

the problem shows ladder’s CM being half-way between the top and the bottom ends — thus

ν = 12 — and the ladder leaning at angle of tangent

tan θ =distance from top end to floor = 4.0 m

distance from bottom end to wall = 3.0 m≈ 1.33. (49)

Plugging these data into eq. (48) and solving for the static friction coefficient µs between the

ladder and the floor, we find

µs × tan θ =Dmax

L×

M

M +m+ ν ×

m

M +m

= 0.70×55.0 kg

(55.0 + 12.0) kg+

1

2×

12.0 kg

(55.0 + 12.0) kg

= 0.664

(50)

and hence

µs =0.664

tan θ≈ 0.50. (51)


To check the stability of a standing body, we need to find the body’s center of gravity (i.e.,

center of mass) and project it vertically down to the ground. The body is stable if this

projection lies within the base supporting the body. If the CM’s projection falls outside the

base of support, the body is unstable and will topple at the first opportunity.

Approximating the Leaning Tower of Pisa as a slightly tilted uniform cylinder, we find its

center of mass at the intersection of the tilted axis of the tower with a horizontal plane at one

11

half of the tower’s height.

CM

x

y

In the coordinate system shown on the above picture, the tower’s axis goes through the points

(x = 0, y = 0) (the center of the tower’s circular base) and (x = L, y = H) at the top of the

tower, where H = 55 m is the tower’s height and L = 4.5 m is the distance by which the top

leans off-center. The center of mass lies between these two points, thus

Xcm = 12L = 2.25 m, Ycm = 1

2H = 27.5 m. (52)

Projecting this center of mass vertically down on the ground, we get a point (denoted by a

blue circle on the above figure) at distance 12L from the center of the tower’s base. The base

is a circle of diameter D = 7.0 m and hence radius R = 12D = 3.5 m. Since the R > 1

2L, the

CM’s projection lies inside the base, so the tower is stable.

If the tower keeps leaning and the displacement of the top increases from L = 4.5 M to

L′ > L, the projection of the tower’s CM onto the ground will be at Xcm = 12L′. As long as

such Xcm is less than the towers radius R, the CM remains strictly above the tower’s base

12

and the tower is stable. But for

Xcm = 12L

′ > R, (53)

the CM’s projection would move outside the tower’s base and the tower would topple. Thus,

the maximal tilt of the tower before it topples is

Lmax = 2× R = D = 7.0 m. (54)

Textbook problem 9.68(a):

The beam lays on two supports A and B, but it isn’t nailed to them. Consequently, the beam

is stable only if the center of gravity of the beam and everything that stands on it lies between

the supports A and B,

XA ≤ Xcm ≤ XB . (55)

The beam itself is uniform, so its own CM lies half-way between its left and right ends,

Xbeamcm =

XC +XD

2. (56)

Consequently, the center of mass of the beam+person system lies at

Xcm =M

M +m×

XC +XD

2+

m

M +m×XP (57)

where M is the beam’s mass, m is the mass of the person standing on the beam, and XP is

that person’s position; in part (a) of the problem, the person stands at the right end of the

beam, hence XP = XD. Thus, the beam remains stable if and only if

XA ≤M

M +m×

XC +XD

2+

m

M +m×XD ≤ XB . (58)

For the beam in question, figure 9-80 of the textbook shows that in the coordinate system

where XC = 0, we have XA = 3.0 m, XB = 15.0 m, and XD = 20.0 m. Plugging these

13

numbers into the inequality (58), we have

3.0 m ≤M

M +m× 10.0 m +

m

M +m× 20.0 m ≤ 15.0 m. (59)

It is easy to see that the first inequality here is true for any m/M ratios, so the beam would

not topple to the left. To make sure it does not topple to the right, we need the second

inequality. Re-expressing it as

(

1 −m

M +m

)

× 10.0 m +m

M +m× 20.0 m ≤ 15.0 m (60)

and hence

m

M +m×

(

20.0 m − 10.0 m)

≤ 15.0 m − 10.0 m, (61)

we solve it in terms of

m

M +m≤

15.0 m − 10.0 m

20.0 m − 10.0 m= 0.50 (62)

and therefore

m ≤ 0.50(M +m) =⇒m

M≤

0.50

1− 0.50= 1.00. (63)

Thus, the beam would remain stable as long as the person standing at the beam’s end is

lighter than the beam itself.

Numerically, we are given the beam’s weight Mg = 550 N, so the person should weigh

less than 550 N. In terms of mass, he or she should have less than 550 N/g ≈ 56 kg or about

124 pounds.

14

⋆ ⋆ ⋆

Textbook question 10.3:

In an open container, the pressure of water depends only on the density. Thus, the force with

which the water presses down on the container’s bottom is

Fb = ρwater × g × depth× Abottom (64)

regardless of the container’s shape. The three containers on figure 10–44 have similar bottom

areas and water depth, so the water presses with the same force (64) on the bottom of each

container.

However, only for the cylindrical container (left) is this force equal to the weight of the

water. The middle container has more water, mg > Fb, while the right container has less

water, mg < Fb.

To resolve this paradox, consider the water pressure on the walls of each container. In the

left container, the walls a vertical, so the forces exerted by water pressure are horizontal. They

do not affect the vertical balance-of-forces equation for the water, so we need Fb −mg = 0 —

which is indeed true for this container.

In the middle container, the walls expand towards the top, so the water force on those

walls has both outward and downward components. By the Third Law, the walls exert an

upward force on the water, hence the vertical balance-of-forces equation becomes

∑

Fy = Fb + Fwy − mg = 0 (65)

where Fwy > 0, and this explains why the bottom’s force Fb is less than the water’s weight

mg.

Finally, the right container has walls narrowing towards the top so the water force ⊥ to

those walls have both outward and upward components. By the Third Law, the wall’s force

15

on the water has a downward components, hence

Fb + Fwy − mg = 0 for Fwy < 0, (66)

which requires Fb > mg: the bottom of the container has to push the water up harder than

the water’s weight to compensate for the downward force from the container walls.

Textbook question 10.6:

In a non-moving fluid of density ρ, the pressure increases with depth as

∆P = const + ρg × depth, (67)

so that pressure difference between the top and bottom of any volume of the fluid would

provide a buoyant force canceling the fluid’s weight. In terms of the vertical coordinate z and

the horizontal coordinates x and y,

P (x, y, z) = const − ρgz regardless of x or y, (68)

and it does not matter if the vessel containing the fluid is straight or crooked.

If we neglect the flow of blood in the arteries and tread it as a fluid at rest, the hydrostatic

equation (68) would tell us that the blood pressure in the patient’s feet should be higher than

in his/her head by

P feet − P head = ρblood × g ×(

zhead − zfeet = patient’s height)

≈ 1050 kg/m3 × 9.8 N/kg × 1.75 m ≈ 18, 000 Pa = 135 mm.Hg.(69)

This is a rather large difference, so what exactly do we mean by the patient’s blood pressure?

Where exactly in the patient’s body do we measure it?

Medically, the blood pressure is particularly important for understanding the condition

of the patients heart, and it’s usually measured using a cuff around the patient’s arm, so by

definition the blood pressure of a patient means the blood pressure in the arteries of his/her

arms at the level of his/her heart.

16

Also, since the blood pressure changes with time according to the patient’s pulse, it is

usually quoted as two numbers: the systolic pressure (maximum) and the diastolic pressure

(minimum), both in units of millimeters of mercury. For example, blood pressure of 140/80

means the systolic pressure of 140 mm.Hg and the diastolic pressure of 80 mm.Hg. (Note that

the slash in 140/80 is not the division operator.) And of course, both the systolic and the

diastolic pressures should be measured at the level of the patient’s heart.

If you measure the blood pressure significantly above or below the patient’s heart, you

measurement will differ by ρg∆z from what the doctor wants to know. For example, if you

raise the patients arm so that the cuff is 40 cm (16 inch) above his head, your measurement

will be off by

∆P = −ρg∆z = −1050 kg/m3 × 9.8 N/kg× 0.40 m ≈ −4100 Pa = −31 mm.Hg. (70)

Thus, if the patient’s true blood pressure (at the heart level) is 140 (systolic) / 80 (diastolic),

your measurement will show only 109 (systolic) / (49) diastolic. On the other hand, if your

measurement reads 160 (systolic) / 95 (diastolic) — which is rather high, but not too bad —

the real blood pressure at the heart level would be much more alarming 191 (systolic) / 126

(diastolic).

To avoid bad medical mistakes like this, blood pressure should be always measured at the

level of the patient’s heart.

PS: For a moving fluid — such as blood in a living person’s arteries — the hydrostatic

equation (68) needs to be corrected to include two effects: First, as the speed of the flow

varies, acceleration or deceleration of the fluid requires a net force, so the buoyant force due

to pressure increasing with depth should balance this force as well as the fluid’s weight. But

the blood flows rather slowly, and its acceleration / deceleration rate is much smaller than g,

so this effect is too small to matter.

The second effect is viscosity, a kind of a velocity-dependent friction between the fluid

and the vessel through which it flows. Because of viscosity, making fluid flow at finite speed

through a thin pipe requires more pressure at the inlet than at the outlet. For the blood, the

viscosity has little effect on the blood pressure in the big arteries (or a healthy person), but

17

it become important for the small arteries and especially for the capillaries. Because of that,

the arterial pressure in the feet would be pretty close to the hydrostatic

P feet = P heart + ρg(zheart − zfeet) ≈ P heart + (100± 10) mm.Hg, (71)

but the pressure in the veins would be much lower. Often, the pressure in the veins of the feet

ends up lower than the 100 mm.Hg, which means that by the time the blood gets back to the

heart, its gauge pressure becomes negative, and the heart need to work very hard to provide

this suction. To ease the situation, leg veins have valves which act as auxiliary pumps when

a person walks or runs.


(a) To lift the car up, the output piston must provide force F2 = mg = 970 kg × 9.8 N/kg =

9500 N. This force comes from the pressure of the liquid on the output piston, thus

F2 = P ×A2 (72)

where P is the gauge pressure of the liquid and A2 is the area of the output piston. In terms

of the pistons diameter D2 = 2R2 = 18 cm,

A2 = π(12D2)2 ≈ 254 cm2, (73)

hence to provide the force F2 lifting the car we need pressure

P =F2

A2=

9500 N

0.0254 m2≈ 373, 000 Pa = 3.69 atm. (74)

By Pascal’s Law, the input piston should exert the same pressure on the liquid. To provide

this pressure under the input force F1 = 250 N, the input piston should have area

A1 =F1

P=

250 N

373, 000 Pa= 6.7 cm2 (75)

and hence diameter

D1 = 2R1 = 2√

A1/π = 2.9 cm. (76)

18

PS: Algebraically,

A1 = A2 ×F1

F2=⇒ D1 = D2 ×

√

F1

F2. (77)

(b) The work done lifting the car by ∆z2 = 12 cm is simply

W = (F2 = mg)×∆z2 = 9500 N× 0.12 m = 1140 J ≈ 1100 J.

(c–d) As the car moves up by ∆z2, the liquid volume under the output piston increases by

∆V = A2 ×∆z2 = 254 cm2 × 12 cm ≈ 3, 050 cm3 = 3.05 L. (78)

The liquid inside the hydraulic lift is approximately incompressible, so to get extra liquid

under the output piston, it the same volume ∆V be squeezed from under the input piston.

Thus,

∆V = A1 ×∆x1 =⇒ ∆x1 =∆V

A1=

3050 cm3

6.7 cm2≈ 456 cm, (79)

i.e. the input piston should be pushed in 4.56 m or about 15 feet!

Such distances are rather impractical for a single stroke of the input piston, so the real

hydraulic lifts have two liquid vessels — a pressurized vessel connected to the output cylinder

and an un-pressurized tank of extra liquid — and a valve switching the input cylinder between

the two vessels. The input piston acts as a pump: during its upward stroke it sucks in the

liquid from the un-pressurized tank, while during the downward stroke it forces it into the

pressurized vessel connected to the output cylinder.

During the each cycle of the input piston, it pumps the volume of liquid

∆V = A1 ×∆x1 = 6.7 cm2 × 13 cm = 87 cm3 (80)

into the pressurized vessel and hence into the output cylinder. Consequently, the car goes up

19

by

∆z2 =∆V

A2=

87 cm3

254 cm2= 0.343 cm. (81)

Thus, to lift the car by 12 cm, the input piston needs to make

N =12 cm

0.343 cm= 35 strokes. (82)

(e) To show that the energy is conserved — or rather that the output work on the car is

equal to the input work on the lift — consider what happens when the input piston goes up

and down. On the upward stroke, it sucks the liquid out of the un-pressurized tank. In the

absence of gauge pressure, this takes no input force on the piston and hence requires no work.

(In practice, it does take a little input force to overcome the friction forces in the piston and

the viscous resistance of the working liquid, but such forces are small and so waste only a

little input work.) On the downward stroke, the liquid is pushed into the pressurized tank.

This requires input force

F1 = P ×A1

and hence consumes input work

Win = F1 ×∆x1 = P ×A1 ×∆x1 = P ×∆V (83)

where ∆V is the volume of the liquid pumped into the pressurized vessel and hence into the

output cylinder. In that cylinder, the output piston goes up by ∆z2 = ∆V/A2 and hence

performs output work

Wout = F2 ×∆z2 = P ×A2 ×∆z2 = P ×∆V. (84)

Note that

Wout = P ×∆V = Win (85)

— the output work equals the input work.

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Numerically, the input work during one stroke of the input piston is

Win(1) = F1 ×∆x1 = 250 N× 13 cm = 32.5 J (86)

so during the N = 35 strokes required to lift the car (see part (d)), the net input work is

W netin = N ×Win(1) = 35× 32.5 J ≈ 1100 J (87)

— which is indeed agrees with the output work found in part (b), Wout ≈ 1100 J.


The gauge pressure in a tire is the difference between the absolute pressure of the air inside

the tire and the atmospheric air pressure on the outside,

P gauge = P abs − P atm. (88)

It’s the gauge pressure that controls the net force of the air on the rubber. For a part of the

tire surface that has area A, the net force is

F net = F from inside out − F from outside in = P abs × A − P atm × A = P gauge ×A. (89)

Most of the tire’s surface is curved, so the tension of the stretched rubber provides an inward

force canceling the outward force (89) of the air pressure. But a small area Ab at the bottom

of the tire is flattened by the contact with the ground, and over this area, there is no net force

due to rubber tension. Instead, the downward force P gauge × Ab is canceled by the normal

force N from the ground,

N − P gauge × Ab = 0. (90)

Since the car has 4 tires, the net normal force from the ground on the car is 4N . This

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force balances the car’s weight Mg. Hence

Mg = 4×N = 4× P gauge × Ab . (91)

For the car in question, the weight is

Mg = 4× 240, 000 Pa× 0.022 m2 = 21, 100 N, (92)

or 4750 pounds. (Which means it’s probably an SUV rather than a car, but what the heck. . . )

Non-textbook problem #2:

The water in a lake is at rest, so the hydrostatic equation tells us that the pressure increases

with depth as

P = const + ρg × depth (93)

where ρ = 1000.0 kg/m3 is the density of water. At the surface of the lake, the water pressure

equals to the air pressure,

P sur = P air = 98 kPa, (94)

so at the bottom of the lake the pressure is

P bot = P air + ρgD (95)

where D is the lake’s depth. Note that both pressures (94) and (95) are absolute pressures

rather than gauge pressures.

According to the Boyle’s Law,⋆volume of a fixed amount of gas held at constant temper-

ature is inversely proportional to the absolute pressure of the gas, hence P × V = const. For

⋆ Historically, several gas laws were discovered between 17 th and 19 th centuries. The oldest gas law —relating the pressure and the volume of a gas — was discovered in 1664 by Robert Boyle in Ireland,hence the name Boyle’s Law. (The same law was independently discovered in 1676 by Edme Mariottein France, so in some countries this law is called the Boyle–Mariotte Law.) Eventually, Boyle’s law wascombined with Charles’s, Gay–Lussac’s, and Avogadro’s laws into the universal gas law (also known asthe ideal gas law). If you have taken a chemistry class, you should already know this law in all its aspects.If not, you can look it up in §13.7 of the Giancoli’s textbook.)

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the bubble in question, this means

P bot × V @bot = P sur × V @sur (96)

and consequently

P bot

P sur=

V @sur

V @bot=

3.0 cm3

1.0 cm3= 3.0. (97)

On the other hand, according to eqs. (94) and (95),

P bot

P sur=

P air + ρgD

P air= 1 +

ρgD

P air, (98)

and combining this formula with eq. (97), we arrive at

1 +ρgD

P air= 3.0. (99)

This is a simple equation for the lake’s depth D, and the solution is

D =P atm

ρg× (3.0 − 1) =

98, 000 N/m2

1000.0 kg/m3 × 9.8 N/kg× (3.0− 1) = 20.0 m. (100)

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PHY–302 K. Solutions for Problem set # 10. Non-textbook...

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