Rotational Statics

Question

A gymnast with mass m_g = 43.0 kg stands on an improvised balance beam that sits on, but is not attached to, two supports. The beam has a mass m_b = 113 kg and length L = 5.00 m. Each support is 1/3 of the beam's length from each end.

Initially, the gymnast stands at the left end of the beam.

Later, the gymnast (not carrying any additional mass) moves to stand directly over the right support.

What is the magnitude F_L of the force that the left support exerts on the beam?

Answer 1

You could make a sketch that might look like this:

With the lower portion having the loads on it. I put the rotational point at the right support.

The W_L is the weight of the left portion of the beam (left of the rotational point). That weight is 2/3 of the whole so it is (2/3)(113)(9.8) = 738.3 N and the distance from the rotational point is 5/3 or 1.667 m so the torque it supplies is (738.3)(1.667) = 1230.4 Nm.

The FL is the force the left support provides to the beam and is what we are solving for. The torque it provides is (FL)((1.667) = 1.6667F_L but is negative since it provides a clockwise torque so -1.6667F_L

Since both FR and WG are at the rotational point, they provide no torque.

WR is the weight of the right portion of the beam and is (1/3)(113)(9.8) = 369.13 N and the distance from the rotational point is half of 1/3 of the beam length or (1/6)(5) = 0.8333 m therefore the torque it supplies is (369.13)(0.8333) = 307.6 Nm but it is negative since it provides a clockwise torque.

Since there is no rotation happening, the sum of the torques is zero:

∑τ = 0 = 1230.4 - 1.6667F_L - 307.6

1.6667F_L = 922.83

F_L = 554 N