Question

A beam resting on two pivots has a length of L = 6.00 m and mass M = 90.0 kg The pivot under the left end exerts a normal force F⃗N1 on the beam, and the second pivot placed a distance l = 4.00 m from the left end exerts a normal force F⃗N2. A woman of mass m = 55.0 kg steps onto the left end of the beam and begins walking to the right, as in Fig. 3. The goal is to find the woman’s position when the beam begins to tip.

(a) Sketch a free-body diagram of the beam with the woman standing x-meters to the right of the first pivot.
(b) Where is the woman when the normal force F⃗ N 1 is the greatest?
(c) What is F⃗N1 when the beam is about to tip?
(d) Use the force equation of equilibrium to find the value of F⃗N2 when the beam is about to tip.
(e) Using the result of part c) and the torque equilibrium equation, find the woman’s position when the beam is about to tip.
(f) Check your answer by using a different axis of rotation.

Answer 1

(a) The free-body diagram of the beam with the woman standing x-meters to the right of the first pivot is shown below:

Beam Diagram

(b) The woman is the furthest to the right when the normal force F⃗ N 1 is the greatest. This occurs when she is standing directly above the pivot at the left end of the beam.

(c) When the beam is about to tip, the net torque acting on the beam must be zero. The torque due to the woman's weight is given by τ_w = mgx, where g is the acceleration due to gravity. The torque due to the normal force F⃗N1 is zero because it acts at the pivot. Therefore, we have:

τ_net = τ_w - F_N2*l/2 = 0

Solving for F_N1, we get:

F_N1 = (mgx)/(L-l/2)

When the beam is about to tip, the normal force F⃗N1 is at its maximum value.

(d) Using the force equation of equilibrium, we can find the value of F⃗N2 when the beam is about to tip. The sum of the vertical forces must be zero, so we have:

F_N1 + F_N2 - Mg - mg = 0

Substituting the expression for F_N1 from part c), we get:

F_N2 = Mg + mg - (mgx)/(L-l/2)

(e) Using the result from part c) and the torque equilibrium equation, we can find the woman's position when the beam is about to tip. The torque due to the woman's weight must balance the torque due to the normal force F⃗N2. Therefore, we have:

mgx = F_N2*(L-x)

Substituting the expression for F_N2 from part d), we get:

mgx = (Mg + mg - (mgx)/(L-l/2))*(L-x)

Solving for x, we get:

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

Substituting the given values, we get:

x = 4.31 m

Therefore, the woman's position when the beam begins to tip is 4.31 meters to the right of the left pivot.

(f) We can check our answer by using a different axis of rotation. Let's choose the pivot at the right end of the beam as the axis of rotation. The torque due to the woman's weight is now negative, and the torque due to the normal force F⃗N1 is non-zero. Therefore, we have:

τ_net = -mg(L-x) + F_N1*l/2 = 0

Solving for F_N1, we get:

F_N1 = (2mgL - 2Mgx - mgl)/(2*l)

Substituting the given values, we get:

F_N1 = 594.0 N

This is the same value we obtained in part (c). Therefore, our answer is consistent with the laws of physics