A gymnast with mass m1 = 44 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 109 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
a. What is the force the left support exerts on the beam?
b. What is the force the right support exerts on the beam?
c. How much extra mass could the gymnast hold before the beam begins to tip?
d. Now the gymnast (not holding any additional mass) walks directly above the right support.
What is the force the left support exerts on the beam?
e. What is the force the right support exerts on the beam?
f. At what location does the gymnast need to stand to maximize the force on the right support? (choose correct answer)
(i) at the center of the beam
(ii) at the right support
(iii) at the right edge of the beam
To solve this problem, we can use the concept of torque and equilibrium.
(a) To find the force the left support exerts on the beam, we need to consider the torques acting on the beam about the right support. The torque due to the gymnast's weight is given by mg * L/3, where mg is the weight of the gymnast and L/3 is the distance between the left support and the center of mass of the beam. The torque due to the beam's weight is m2 * g * (2L/3), where m2 is the mass of the beam and 2L/3 is the distance between the right support and the center of mass of the beam.
Since the beam is in equilibrium, the sum of these torques must be zero. Therefore, we have:
mg * L/3 + m2 * g * (2L/3) = 0
Substituting the given values, m1 = 44 kg, m2 = 109 kg, and L = 5 m, we can solve for the force the left support exerts on the beam.
(b) Similarly, to find the force the right support exerts on the beam, we consider torques about the left support. The torque due to the gymnast's weight is mg * (2L/3), and the torque due to the beam's weight is m2 * g * L/3. Again, since the beam is in equilibrium, we have:
mg * (2L/3) + m2 * g * L/3 = 0
Substituting the given values, we can solve for the force the right support exerts on the beam.
(c) To find the maximum extra mass the gymnast can hold before the beam begins to tip, we need to consider the torque required to balance the torques due to the gymnast and the beam's weights. The maximum torque the gymnast can exert will be when the gymnast is at the right edge of the beam. The torque due to the gymnast's weight is mg * L, and the torque due to the beam's weight is m2 * g * L/3. Again, since the beam is in equilibrium, we have:
mg * L + m2 * g * L/3 = 0
Substituting the given values, we can solve for the maximum extra mass the gymnast can hold.
(d) When the gymnast walks directly above the right support, the torque due to the gymnast's weight becomes zero, as it acts directly through the axis of rotation. Therefore, the torque equation for the left support is:
m2 * g * (2L/3) = 0
(e) Similar to part (b), we can use the torque equation about the left support to find the force the right support exerts on the beam.
(f) To maximize the force on the right support, the gymnast should stand at the center of the beam (i). This is because the torque due to the gymnast's weight is largest when the distance between the gymnast and the center of mass of the beam is the largest. When the gymnast stands at the center, the torque due to the gymnast's weight is zero, and the full torque due to the beam's weight is exerted on the right support.