fisica1-youn-e-freedman-exercicios-resolvidos

11.8: The weight of the motor is 400 N + 600 N − 200 N = 800 N. Of the myriad ways
to do this problem, a sneaky way is to say that the lifters each exert 100 N to the lift the
board, leaving 500 N and 300 N to the lift the motor. Then, the distance of the motor
(2.00 m)(300 N)
from the end where the 600-N force is applied is = 0.75 m .The center of
gravity is located at
is applied.
(200 N)(1.0
m) + (800
(1000 N)
N)(0.75 m)
=
0.80 m
(800 N)
from the end where the
600 N force
Lw
11.9: The torque due to T is − T h = − cotθ
h,
and the torque due to T is T D Lw .
x
x
D
y y
=
h
The sum of these torques is Lw( 1− cotθ).
D
From Figure (11.9(b)), h = D tanθ,
so the net
torque due to the tension in the tendon is zero.
11.10: a) Since the wall is frictionless, the only vertical forces are the weights of the
man and the ladder, and the normal force. For the vertical forces to balance,
n
2
= w1
+ wm
= 160 N + 740 N = 900 N, and the maximum frictional forces is
µ
s
n2
= (0.40)(900 N) = 360 N (see Figure 11.7(b)). b) Note that the ladder makes contact
with the wall at a height of 4.0 m above the ground. Balancing torques about the point of
contact with the ground,
( 4.0 m) n
1
= (1.5 m)(160 N) + (1.0 m)(3 5))(740 N) = 684 N ⋅ m,
so n = 171.0
1
N, keeping extra figures. This horizontal force about must be balanced by
the frictional force, which must then be 170 N to two figures. c) Setting the frictional
force, and hence n
1, equal to the maximum of 360 N and solving for the distance x along
the ladder,
( 4.0 m)(360 N) = (1.50 m)(160 N) + x(3
5)(740 N),
so x = 2.70 m, or 2.7 m to two figures.
11.11: Take torques about the left end of the board in Figure (11.21). a) The force F at
the support point is found from
F( 1.00 m) = + (280 N)(1.50 m) + (500 N)(3.00 m), or F = 1920 N. b) The net force must be
zero, so the force at the left end is ( 1920 N) − (500 N) − (280 N) = 1140 N, downward.