A beam resting on two pivots has a length of L = 6.00 m and mass M = 76.0 kg. Th

Question

A beam resting on two pivots has a length of L = 6.00 m and mass M = 76.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 53.5 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.

a) Where is the woman when the normal force n1 is the greatest?

b) What is n1when the beam is about to tip?

c) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip.

d) Using the result of part (b) 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

e) Check the answer to part (d) by computing torques around the first pivot point.

Explanation / Answer

a)

n1 is maximum when the woman stands at the left end

b)

n1 =0 since the left end looses contact with the pivot when beam starts to tip

c)

When beam is about to tip

weight of beam + weight of woman = n2

Mg + mg = n1

d)

Let the woman is at distance 'r' from second pivot

Taking equilibrium of torque about second Point

mg(r) = Mg (l - L/2)

53.5 r = 76 (4 - 6/2)

r = 1.421

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