A gymnast with mass m 1 = 40 kg is on a balance beam that sits on (but is not at

Question

A gymnast with mass m1 = 40 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 104 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? (center of beam, at the right support, at te right edge of the beam)

Explanation / Answer

since the system at rest

Fnet = 0

FL + FR - m1g - m2g = 0

and it is not rotationg so torque = 0

FR*L/6 - L*FL/6 + m1g*L/2

part 1 )

for left force

FL = (4m1 + m2)*g/2

FL = 1293.6 N

part 2 )

FR = (m2 - 2m1)*g/2

FR = 117.6 N

part 3 )

the beam start to tip when FR = 0

0 = (m2 - 2m1')*g/2

m' = m2/2

dm = extra mass

dm = m1' - m1 = m2/2 - m1

dm = 12 kg

part 4 )

now gymnast start to move

Fnet = 0

FL' + FR' - m1g - m2g = 0

torque = 0

FR'*L/6 - L*FL'/6 - m1g*L/6

FL' = (2m1+m2)*g/2

FL' = 901.6 N

part 5 )

FR' = m2g/2

FR' = 509.6 N

part 6 )

at the rigth edge of the beam

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