A gymnast with mass m 1 = 47 kg is on a balance beam that sits on (but is not at

Question

A gymnast with mass m1 = 47 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 121 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?

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?

at the center of the beam

at the right support

at the right edge of the beam

Explanation / Answer

1) As the beam is in equilibrium, net force and net torque acting on the beam must be equal to zero.

Apply, Net torque about right support = 0

FL*(L/3) - m2*g*(L/2 - L/3) - m1*2*L/3 = 0

FL*L/3 - m2*g*L/6 - m1*2*L/3 = 0

FL/3 = m2*g/6 + 2*m1*g/3

FL = m2*g/2 + 2*m1*g

= 121*9.8/2 + 2*47*9.8

= 1514.1 N

2)

Apply, Net torque about left support = 0

m1*g*L/3 + FR*(L/3) - m2*g*L/6 = 0

FR/3 = -m1*g/3 + m2*g/6

FR = -m1*g + m2*g/2

= -47*9.8 + 121*9.8/2

= 1053.5 N

3) Let m is the required mass.

Apply,

(m1 + m)*g*L/3 = m2*g*L/6

m1 + m = m2/2

m = m2/2 - m1

= 121/2 - 47

= 13.5 kg

4) Apply, net torque about right support = 0

m2*g*L/6 - FL*L/3 = 0

FL/3 = m2*g/6

FL = m2*g/2

= 121*9.8/2

= 592.9 N

5) Apply, net torque about left support = 0

FR*L/3 - m1*g*L/3 - m2*g*L/6 = 0

FR/3 = m1*g/3 + m2*g/6

FR = m1*g + m2*g/2

= 47*9.8 + 121*9.8/2

= 1053.5 N

6) at the right edge of the beam

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