fisica1-youn-e-freedman-exercicios-resolvidos

11.1: Take the origin to be at the center of the small ball; then,
(1.00 kg)(0) + (2.00 kg)(0.580 m)
x
cm
=
= 0.387 m
3.00 kg
from the center of the small ball.
11.2: The calculation of Exercise 11.1 becomes
(1.00 kg)(0) + (1.50 kg)(0.280 m) + (2.00 kg)(0.580 m)
x =
4.50 kg
This result is smaller than the one obtained in Exercise 11.1.
cm
=
0.351m
11.3: In the notation of Example 11.1, take the origin to be the point S , and let the
child’s distance from this point be x . Then,
M ( − D 2) + mx MD
s
cm
=
= 0, x = = 1.125 m,
M + m
2 m
which is ( L 2 − D 2) 2,
halfway between the point S and the end of the plank.
11.4: a) The force is applied at the center of mass, so the applied force must have the
same magnitude as the weight of the door, or 300 N. In this case, the hinge exerts no
force.
b) With respect to the hinge, the moment arm of the applied force is twice the
distance to the center of mass, so the force has half the magnitude of the weight, or
150 N . The hinge supplies an upward force of 300 N − 150 N = 150 N.
11.5: F( 8.0 m)sin 40° = (2800 N)(10.0 m), so F = 5.45 kN, keeping an extra figure.
11.6: The other person lifts with a force of 160 N − 60 N = 100 N. Taking torques
about the point where the 60 - N force is applied,
⎛160
N ⎞
( 100 N) x = (160 N)(1.50 m), or x = (1.50 m)
⎜ = 2.40 m.
100 N
⎟
⎝ ⎠
11.7: If the board is taken to be massless, the weight of the motor is the sum of the
(2.00 m)(600 N)
applied forces, 1000 N. The motor is a distance = 1.200 m from the end where
the 400-N force is applied.
(1000 N)