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A person is pushing a wheelbarrow along a ramp that makes an angle 38 degrees with...

A person is pushing a wheelbarrow along a ramp that makes an angle 38 degrees with the horizontal. The wheelbarrow and load have a combined mass of 28.4kg with the center of mass at the midpoint of the length L. What is the magnitude of the net force that the person must apply in order to push the wheelbarrow up the ramp at a constant velocity, while keeping the wheelbarrow in a level, horizontal orientation? Assume that the radius of the wheel is small enough to ignore. Use g=9.81 m^2/s^2. Please show free body diagram and torques.

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Answer #1

Answer:

(Combined mass)(g) = Total downward force = 28.4 * 9.81 = 278.60 N

X-component force: We're in dynamic equilibrium, so upward force on each end is

(total downward force) / 2 = (278.60) / 2 = 139.3 N

y-component force: Doing force vector and trig magic, we find that the competing horizontal forces on each side is

(tan38)(139.3) = 108.833 N


vector sum: Use the pythagorean theorem [(139.3^2)+(108.833^2)]^1/2 = 176.8 N (Answer)

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