A uniform beam (mass M = 68.0 kg and length L = 5.30 m) is supported by two tria

Question

A uniform beam (mass M = 68.0 kg and length L = 5.30 m) is supported by two triangular supports that are placed a distance D = 3.18 m apart, as shown below. The left-most support is at the very end of the beam. A person (mass m = 29.0 kg) starts walking at constant speed v along the beam, beginning from some point between the two supports.

How close can the person get to the right-most end of the beam without the beam flipping over? (Report your answer as a distance measured from the right-most end of the beam.)
m

Explanation / Answer

Let R1 and R2 are reaction forces from left and right posts respectively

Then, Mg +mg = R1 + R2

R1+ R2 = (68)(9.81) + (29)(9.81) = 951.57 N

When person is at a position x from rightmost end, then torque about leftmost end is zero

mg(L-x) + Mg(L/2) = R2(D)

(29)(9.81)(5.30-x) + (68)(9.81)(5.30/2) = R2(3.18)

1507.797 - 284.49x + 1767.762 = 3.18R2

3275.559 -284.49x = 3.18R2

R2 = 1030.05 - 89.46x

Since R1 + R2 = 951.57 N, R2 must be less than 951.57N

1030.05 - 89.46x < 951.57

x< 0.877 m

Person can be at 0.877 m or more from rightmost end but not less than that

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