A gymnast with mass m 1 = 44 kg is on a balance beam that sits on (but is not at
ID: 1790241 • Letter: A
Question
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 = 110 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.
1) What is the force the left support exerts on the beam?
2) What is the force the right support exerts on the beam?
3) How much extra mass could the gymnast hold before the beam begins to tip?
4)
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?
5) What is the force the right support exerts on the beam?
6) At what location does the gymnast need to stand to maximize the force on the right support?
a) at the center of the beam
b) at the right support
c) at the right edge of the beam
Explanation / Answer
1)
net torque about the right support = 0
m1*g*2L/3 + m2*g*(2L/3 - L/2) - FL*(L/3) = 0
(44*9.8*10/3) + (110*9.8*(10/3 - 5/2)) - FL*5/3 = 0
FL = 1401.4 N <<<<-------ANSWER
===========================
2)
In static equilibrium Fnet = 0
FL + Fr - (m1+m2)*g = 0
1401.4 + Fr - (44+110)*9.8 = 0
force exerted by right support = Fr = 107.8 N
=============================
3)
If the beam begins to tip
Fr = 0
net torque about left = 0
(m1+m)*g*L/3 - m2*g*(L/2 - L/3) = 0
(44+m)*9.8*5/3 - 110*9.8*(5/2 - 5/3) = 0
m = 11 kg
========================
4)
In rotational equilibrium net torque = 0
net torque about right support = 0
FL*L/3 - m2*g*(L/2 - L/3) = 0
FL*5/3 - 110*9.8*(5/2 - 5/3) = 0
Fl = 539 N
-------------------------
5)
along vertical
Fnet = 0
FL + FR - (m1+m2)*g = 0
539 + FR - (44+110)*9.8 = 0
FR = 970 N
==========================
6)
at the right edge of beam
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