Imagine you are standing upright and bolding a 7.0-kg metal pole. The pole is 5.
ID: 1546216 • Letter: I
Question
Imagine you are standing upright and bolding a 7.0-kg metal pole. The pole is 5.0 meters long, and you are holding it horizontally. Assume that you are not applying any horizontal forces to the pole with either hand. Suppose one of your hands (hand 1) is gripping the pole at a distance 2.1 meters from the left end of the pole, and your other hand (hand 2) is gripping the pole at a distance of 1.8 meters from the right end of the pole, meaning your hands are 1.1 meters apart. Following the standard steps for a statics problem, find the magnitude and direction of the force exerted on the pole by each of your hands. Now suppose that one hand (hand 1) is gripping the pole at the left end of the pole, and your other hand (and is gripping the pole at a distance of 1.1 meters from the left end of the pole, Your hands are still 1.1 meters apart. Following the standard steps for a statics problem, find the magnitude and direction of the force exerted on the pole by each of your hands.Explanation / Answer
in case of equilibrium, net torque is 0
net torque about left end
7 * 9.8 * 2.5 - F1 * 2.1 - F2 * (2.1 + 1.1) = 0
net torque about mid point
F1 * (2.5 - 2.1) - F2 * (2.5 - 1.8) = 0
on solving above two equations we'll get
F1 = 43.65 N
F2 = 24.94 N
a) force applied by hand 1 = 43.65 N upward direction
a) force applied by hand 2 = 24.94 N upward direction
in second case
net torque about left end = 0
F2 * 1.1 - 7 * 9.8 * 2.5 = 0
F2 = 155.9 N
torque about mid point = 0
F1 * 2.5 - F2 * (2.5 - 1.1) = 0
155.9 * 2.5 - F2 * (2.5 - 1.1) = 0
F2 = 278.39 N
b) force applied by hand 1 = 155.9 N in downward direction
b) force applied by hand 2 = 278.39 N in upward direction
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