Question

A uniform beam resting on two pivots has a length L = 6.00 m and weight M = 220 lbs. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot, located a distance l = 4.00 m from the end exerts a normal force n2. A woman with a weight of 130 lbs steps onto the left end of the beam and begins walking to the right. How far can the woman walk before the beam tips?

Answer 1

Answer:

x = 4,138 m

Explanation:

For this exercise, let's use the rotational equilibrium equation.

Let's fix our frame of reference on the left side of the pivot, the positive direction for anti-clockwise rotation

         ∑ τ = 0

         n₁ 0 - W L / 2 + n₂ 4 - W_woman  x = 0

         x = (- W L / 2 + 4n2) / W_woman

Let's reduce the magnitudes to the SI System

         M = 6 lbs (1 kg / 2.2 lb) = 2.72 kg

         M_woman = 130 lbs = 59.09 kg

Let's write the transnational equilibrium equation

         n₁ + n₂ - W - W_woman = 0

         n₁ + n₂ = W + W_woman

        n₁ + n₂ = (2.72 + 59.09) 9.8

At the point where the system begins to rotate, pivot 1 has no force on it, so its relation must be zero (n₁ = 0)

          n₂ = 605,738 N

 

Let's calculate

         x = (-2.72 9.8 6/2 + 4 605.738) / 59.09 9.8

         x = 4,138 m