A uniform 240-g meter stick can be balanced by a 240-g weight placed at the 80-c

Question

A uniform 240-g meter stick can be balanced by a 240-g weight placed at the 80-cm mark if the fulcrum is placed at the point marked: 75 cm 70 cm 65 cm 60 cm 55 cm Calculate the torque (magnitude and direction) about the post B of a diving board, exerted by a 54-kg person who is 3 meters from the supporting post B: 530 N m, clockwise. 1590 N m, clockwise. 1590 N m, counterclockwise. 2120 N m, counterclockwise. 2120 N m, clockwise. An object at the surface of Earth (at a distance R from the center of Earth) weighs 270 N. Its weight at a distance 3R from the center of Earth is 10N 30 N 90 N 270 N 810 N A force F = (-5.0 N)i + (7.0 N)j + (-3.0 N)k acts on a particle located

Explanation / Answer

9)

For a meter scale the Center of mass lies at 50 cm

240g(50-x) = 240g(x-80)

2x = 130

x = 65 cm ANSWER IS C

10)

TORQUE ,T =F*R = MG*R = 54*9.8*3 = 1587.6 Nm ~ 1590 Nm

answer is B, direction is in direction of r cross F which turns out to be clockwise

11)

w1 r1^2 = w2 r2^2

270*R^2 = W*9R^2

w = 30 N

answer is B

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