fisica1-youn-e-freedman-exercicios-resolvidos

11.72: a) Take torques about the upper corner of the curb. The force F r acts at a
perpendicular distance R − h and the weight acts at a perpendicular distance
R
2
−
2
2
( R − h) = 2Rh
− h .
Setting the torques equal for the minimum necessary force,
2
2Rh
− h
F = mg .
R − h
b) The torque due to gravity is the same, but the force F r acts at a perpendicular
distance 2R − h,
so the minimum force is ( mg)
2Rh
− hv /2R–h. c) Less force is
required when the force is applied at the top of the wheel.
11.73: a) There are several ways to find the tension. Taking torques about point B (the
force of the hinge at A is given as being vertical, and exerts no torque about B), the
2
2
tension acts at distance r = ( 4.00 m) + (2.00 m) = 4. 47 m and at an angle of
⎛ 2.00 ⎞
φ = 30 ° + arctan ⎜ ⎟ = 56.6°
. Setting
⎝ 4.00 ⎠
Tr sin φ = (500N)(2.00 m) and solving for T givesT
= 268 N . b) The hinge at A is given
as exerting no horizontal force, so taking torques about point D , the lever arm for the
vertical force at point B is (2.00 m) + (4.00 m) tan 30.0°
= 4.31m,
so the horizontal force
(500 N)(2.00 m)
at B is = 232 N. Using the result of part (a),
4.31m
however, ( 268 N)cos 30.0° = 232 N In fact, finding the horizontal force at B first
simplifies the calculation of the tension slightly. c) ( 500 N) − (268 N)sin30.0°
= 366 N.
Equivalently, the result of part (b) could be used, taking torques about point C , to get the
same result.