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Homework #3
12.
Two forces are applied to a car in an effort to
move it, as shown in Figure P4.12. (a) What is the
resultant of these two forces? (b) If the car has a mass
of 3 000 kg, what acceleration does it have? Ignore friction.
14.
The force exerted by the wind on the sails of a
sailboat is 390 N north. The water exerts a force of
180 N east. If the boat (including its crew) has a mass
of 270 kg, what are the magnitude and direction of its
acceleration?
Figure P4.12
15.
Find the tension in each cable supporting the 600-N cat burglar in Figure
P4.15.
Figure P4.15
Figure P4.16
16.
Find the tension in the two wires that support the 100-N light fixture in
Figure P4.16.
17.
A 150-N bird feeder is supported by three cables as
shown in Figure P4.17. Find the tension in each cable.
18.
The leg and cast in Figure P4.18 weigh 220 N (w1).
Determine the weight w2 and the angle α needed so that no
force is exerted on the hip joint by the leg plus the cast.
Figure P4.17
Figure P4.18
21.
The distance between two telephone poles is 50.0 m. When a 1.00-kg bird
lands on the telephone wire midway between the poles, the wire sags 0.200 m.
Draw a free-body diagram of the bird. How much tension does the bird produce
in the wire? Ignore the weight of the wire.
Homework #3 Solutions
4.12
(a) Choose the positive y-axis in the forward direction. We
resolve the forces into their components as
Force
400 N
450 N
Resultant
x-component
200 N
–78.1 N
y-component
346 N
443 N
Fx  122 N
Fy  790 N
The magnitude and direction of the resultant force is
FR 
 Fx 2   Fy 
2
 F 
 799 N ,   tan1  x   8.77 to right of y-axis.
 Fy 


Thus, FR  799 N at8.77 to the rightofthe forw ard direction
(b) The acceleration is in the same direction as FR and has magnitude
a
4.14
FR
799 N

 0.266 m s2
m 3000 kg
Since the two forces are perpendicular to each other, their resultant is:
FR 
180 N 2   390 N 2  430 N , at
  tan1 
390 N 
  65.2 N ofE
 180 N 
Thus, a 
FR 430 N

 1.59 m s2
m 270 kg
a  1.59 m s2 at65.2 N ofE
or
4.15
Since the burglar is held in equilibrium, the tension in the
vertical cable equals the burglar’s weight of 600 N
Now, consider the junction in the three cables:
Fy  0 , giving T2 sin 37.0  600 N  0
or
T2  997 N in the inclined cable
Also, Fx  0 which yields T2 cos37.0  T1  0
or
4.16
T1   997 N  cos37.0  796 N in the horizontalcable
From Fx  0 , T1 cos40.0  T2 cos40.0  0
or
T1  T2
Then, Fy  0 gives 2 T1 sin 40.0  100 N  0
yielding T1  T2  77.8 N
4.17
From Fx  0 , T1 cos30.0  T2 cos60.0  0
or
T2  1.73 T1
(1)
Then Fy  0 becomes
T1 sin30.0  1.73 T1  sin60.0  150 N  0
which gives T1  75.0 N in the rightside cable
Finally, Equation (1) above gives T2  130 N in the leftside cable
4.18
If the hip exerts no force on the leg, the system must be
in equilibrium with the three forces shown in the freebody diagram.
Thus Fx  0 becomes
w 2 cos  110 N  cos40
(1)
From Fy  0 , we find
w 2 sin  220 N  110 N  sin40
(2)
Dividing Equation (2) by Equation (1) yields
 220 N   110 N  sin 40 
  61
110 N  cos40


  tan1 
Then, from either Equation (1) or (2), w 2  1.7  102 N
4.21
m  1.00 kg and m g  9.80 N
 0.200 m 
 0.458
 25.0 m 
  tan1 
Require that Fy  0 ,
2T sin   m g
T
9.80 N
 613 N
2sin