Home Work 12 – Chapter 12

2. Figure 12-14 shows three situations in which the same horizontal rod is supported by a hinge on a wall

at one end and a cord at its other end. Without written calculation, rank the situations according to the

magnitudes of (a) the force on the rod from the cord, (b) the vertical force on the rod from the hinge,

and (c) the horizontal force on the rod from the hinge, greatest first.

Figure 12-14 Question 1.

Solutions

(a) 1 and 3 tie, then 2; (b) all tie; (c) 1 and 3 tie, then 2 (zero)

1. Because g varies so little over the extent of most structures, any structure's center of gravity

effectively coincides with its center of mass. Here is a fictitious example where g varies more

significantly. Figure 12-23 shows an array of six particles, each with mass m, fixed to the edge of a

rigid structure of negligible mass. The distance between adjacent particles along the edge is 2.00 m.

The following table gives the value of g (m/s2) at each particle's location. Using the coordinate system

shown, find (a) the x coordinate xcom and (b) the y coordinate ycom of the center of mass of the six-

particle system. Then find (c) the x coordinate xcog and (d) the y coordinate ycog of the center of gravity

of the six-particle system.

Particle g

1 8.00

2 7.80

3 7.60

4 7.40

5 7.60

6 7.80

Figure 12-23 Problem 1

Solutions

(a) The center of mass is given by

com

0 0 0 ( )(2.00 m) ( )(2.00 m) ( )(2.00 m)1.00 m.

6

m m mx

m

(b) Similarly, we have

com

0 ( )(2.00 m) ( )(4.00 m) ( )(4.00 m) ( )(2.00 m) 02.00 m.

6

m m m my

m

(c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this

problem, we have

6

1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6cog 6

1 1 2 2 3 3 4 4 5 5 6 6

1

0.987 m.i i i

i

i i

i

x m gx m g x m g x m g x m g x m g x m g

xm g m g m g m g m g m g

m g

(d) Similarly, we have

6

1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6cog 6

1 1 2 2 3 3 4 4 5 5 6 6

1

0 (2.00)(7.80 ) (4.00)(7.60 ) (4.00)(7.40 ) (2.00)(7.60 ) 0

8.0 7.80 7.60 7.40 7.60 7.80

1.97 m.

i i i

i

i i

i

y m gy m g y m g y m g y m g y m g y m g

ym g m g m g m g m g m g

m g

m m m m

m m m m m m

10. The system in Fig. 12-26 is in equilibrium, with the string in the center exactly horizontal.

Block A weighs 40 N, block B weighs 50 N, and angle is 35°. Find (a) tension T1, (b) tension T2, (c)

tension T3, and (d) angle θ.

Figure 12-26 Problem 10

Solutions

(a) Analyzing vertical forces where string 1 and string 2 meet, we find

1

40N49N.

cos cos 35

AwT

(b) Looking at the horizontal forces at that point leads to

2 1 sin35 (49N)sin35 28N.T T

(c) We denote the components of T3 as Tx (rightward) and Ty (upward). Analyzing horizontal forces

where string 2 and string 3 meet, we find Tx = T2 = 28 N. From the vertical forces there, we conclude Ty =

wB = 50 N. Therefore,

2 2

3 57 N.x yT T T

(d) The angle of string 3 (measured from vertical) is

1 1 28tan tan 29 .

50

x

y

T

T

11. Figure 12-27 shows a diver of weight 580 N standing at the end of a diving board with a length of L =

4.5 m and negligible mass. The board is fixed to two pedestals (supports) that are separated by

distance d = 1.5 m. Of the forces acting on the board, what are the (a) magnitude and (b) direction (up

or down) of the force from the left pedestal and the (c) magnitude and (d) direction (up or down) of

the force from the right pedestal? (e) Which pedestal (left or right) is being stretched, and (f) which

pedestal is being compressed?

Figure 12-27 Problem 11.

Solutions

We take the force of the left pedestal to be F1 at x = 0, where the x axis is along the diving board. We

take the force of the right pedestal to be F2 and denote its position as x = d. W is the weight of the diver,

located at x = L. The following two equations result from setting the sum of forces equal to zero (with

upward positive), and the sum of torques (about x2) equal to zero:

1 2

1

0

( ) 0

F F W

F d W L d

(a) The second equation gives

1

3.0m(580 N) 1160 N

1.5m

L dF W

d

which should be rounded off to 3

1 1.2 10 NF . Thus, 3

1| | 1.2 10 N.F

(b) F1 is negative, indicating that this force is downward.

(c) The first equation gives 2 1 580N 1160N 1740NF W F

which should be rounded off to 3

2 1.7 10 NF . Thus, 3

2| | 1.7 10 N.F

(d) The result is positive, indicating that this force is upward.

(e) The force of the diving board on the left pedestal is upward (opposite to the force of the pedestal on

the diving board), so this pedestal is being stretched.

(f) The force of the diving board on the right pedestal is downward, so this pedestal is being

compressed.

22. In Fig. 12-37, a 55 kg rock climber is in a lie-back climb along a fissure, with hands pulling on one

side of the fissure and feet pressed against the opposite side. The fissure has width w = 0.20 m, and the

center of mass of the climber is a horizontal distance d = 0.40 m from the fissure. The coefficient of

static friction between hands and rock is μ1 = 0.40, and between boots and rock it is μ2 = 1.2. (a) What

is the least horizontal pull by the hands and push by the feet that will keep the climber stable? (b) For

the horizontal pull of (a), what must be the vertical distance h between hands and feet? If the climber

encounters wet rock, so that μ1 and μ2 are reduced, what happens to (c) the answer to (a) and (d) the

answer to (b)?

Figure 12-37 Problem 22.

Solutions

(a) The problem asks for the person’s pull (his force exerted on the rock) but since we are examining

forces and torques on the person, we solve for the reaction force 1NF (exerted leftward on the hands by

the rock). At that point, there is also an upward force of static friction on his hands, f1, which we will

take to be at its maximum value 1 1NF . We note that equilibrium of horizontal forces requires

1 2N NF F (the force exerted leftward on his feet); on his feet there is also an upward static friction

force of magnitude 2FN2. Equilibrium of vertical forces gives

2

1 2 1

1 2

+ = 0 = = 3.4 10 N.+

N

mgf f mg F

(b) Computing torques about the point where his feet come in contact with the rock, we find

1 1

1 1

1

++ = 0 = = 0.88 m.

N

N

N

mg d w F wmg d w f w F h h

F

(c) Both intuitively and mathematically (since both coefficients are in the denominator) we see from part

(a) that 1NF would increase in such a case.

(d) As for part (b), it helps to plug part (a) into part (b) and simplify:

( )

from which it becomes apparent that h should decrease if the coefficients decrease.

24. In Fig. 12-39, a climber with a weight of 533.8 N is held by a belay rope connected to her climbing

harness and belay device; the force of the rope on her has a line of action through her center of mass.

The indicated angles are θ = 40.0° and = 30.0°. If her feet are on the verge of sliding on the vertical

wall, what is the coefficient of static friction between her climbing shoes and the wall?

Figure 12-39 Problem 24.

Solutions

As shown in the free-body diagram, the forces on the climber consist of T from the rope, normal force

NF on her feet, upward static frictional force ,sf and downward gravitational force mg .

Since the climber is in static equilibrium, the net force acting on her is zero. Applying Newton’s second

law to the vertical and horizontal directions, we have

net,

net,

0 sin

0 cos .

x N

y s

F F T

F T f mg

In addition, the net torque about O (contact point between her feet and the wall) must also

vanish:

net0 sin sin(180 )O

mgL TL

From the torque equation, we obtain

sin / sin(180 ).T mg

Substituting the expression into the force equations, and noting that s s Nf F , we find the coefficient

of static friction to be

cos sin cos / sin(180 )

sin sin sin / sin(180 )

1 sin cos / sin(180 ).

sin sin / sin(180 )

ss

N

f mg T mg mg

F T mg

With 40 and 30 , the result is

1 sin cos / sin(180 ) 1 sin 40 cos30 / sin(180 40 30 )

sin sin / sin(180 ) sin 40 sin 30 / sin(180 40 30 )

1.19.

s

28. In Fig. 12-43, suppose the length L of the uniform bar is 3.00 m and its weight is 200 N. Also, let the

block's weight W = 300 N and the angle θ = 30.0°. The wire can withstand a maximum tension of

500 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at

this maximum x, what are the (b) horizontal and (c) vertical components of the force on the bar from

the hinge at A?

Figure 12-43 Problems 28.

Solutions

(a) Computing torques about point A, we find

(

)

We solve for the maximum distance:

maxmax

sin / 2 (500 N)sin30.0 (200 N) / 23.00 m 1.50m.

300 N

bT Wx L

W

(b) Equilibrium of horizontal forces gives max= cos = 433N.xF T

(c) And equilibrium of vertical forces gives max= + sin = 250N.y bF W W T