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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