fisica1-youn-e-freedman-exercicios-resolvidos

11.43: For the airplane to remain in level flight, both ∑ F = 0 and ∑τ
= 0 .
Taking the clockwise direction as positive, and taking torques about the center of mass,
Forces: − F −W
+ F 0
tail wing
=
Torques: − 3.66 m) F + (.3 m) F 0
(
tail wing
=
A shortcut method is to write a second torque equation for torques about the tail, and
solve for the F −(3.66 m)(6700 N) + (3.36 m) F 0.
This gives
F
wing
wing
:
wing
=
= 7300 N(up), and Ftail
= 6700 N − 7300 N = −600 N(down).
Note that the rear stabilizer provides a downward force, does not hold up the tail of the
aircraft, but serves to counter the torque produced by the wing. Thus balance, along with
weight, is a crucial factor in airplane loading.
11.44: The simplest way to do this is to consider the changes in the forces due to the
extra weight of the box. Taking torques about the rear axle, the force on the front wheels
1.00 m
is decreased by 3600 N = 1200 N,
3.00 m
so the net force on the front wheels
3
is10,780 N −1200 N = 9.58×
10 N to three figures. The weight added to the rear wheels
is then 3600 N + 1200 N = 4800 N, so the net force on the rear wheels is
4
8820 N + 4800 N = 1.36 × 10 N, again to three figures.
b) Now we want a shift of 10,780 N away from the front axle. Therefore,
m
W 1.00
= 10,780 N and so w = 32,340 N.
3.00 m
11.45: Take torques about the pivot point, which is 2.20 m from Karen and 1.65 m
from Elwood. Then wElwood (1.65 m) = (420 N)(2.20 m) + (240 N)(0.20 m),
so Elwood
weighs 589 N. b) Equilibrium is neutral.