A beam resting on two pivots has a length of L = 6.00 m and mass M = 94.0 kg. Th

Question

A beam resting on two pivots has a length of L = 6.00 m and mass M = 94.0 kg. The pivot under the left end exerts a normal force n_1 on the beam, and the second pivot placed a distance l = 4.00 m from the left end exerts a normal force n_2. A woman of mass m = 52.0 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman x meters to the right of the first pivot, which is the origin. (Submit a file with a maximum size of 1 MB.) Where is the woman when the normal force n_1 is the greatest? What is n_1 when the beam is about to tip? Use the force equation of equilibrium to find the value of n_2 when the beam is about to tip. Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. Check the answer to part (e) by computing torques around the first pivot point.

Explanation / Answer

in equilibrium

n1 + n2 = W1 + W2

0 + n2 = (94*9.8)+(52*9.8)

n2 = 1430.8 N

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e)

torque about the second pivot = 0

M*g*(l - L/2 ) = m*g*(x-l)

94*9.8*(4 - 6/2) = 52*9.8*(x-4)

x = 5.81 m <<--answer

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f)

torque about the left end = 0

n2*l = M*g*L/2 + m*g*x

1430.8*4 = (94*9.8*3) + (52*9.8*x)
x = 5.81 m

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