A gymnast with mass m 1 = 43 kg is on a balance beam that sits on (but is not at

Question

A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 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.

4) 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

part 1 )

force upward = force downward

(S1) + (S2) = (m1)(g) + (m2)(g)

(S1) + (S2) = (43 kg)(9.8 ) + (115 kg)(9.8)

(S1) + (S2) = 1548.4 N [1]

torques balance (using the right end as a reference):

(S1)(2/3)(L) + (S2)(1/3)(L) = (m1)(g)(L) + (m2)(g)(1/2)(L)

(S1)(2) + (S2) = 3[(m1)+ (m2)/2](g)

(S1)(2) + (S2) = 3[(43 kg)+ (115 kg)/2](9.8)

(S1)(2) + (S2) = 2954.7 N [2]

1)

[2] - [1]:

(S1) = 2954.7 - 1548.4 N

(S1) = 1406.3 N

2)

1406.3 N + (S2) = 1548.4 N

(S2) = 142.1 N

3)

Using the left support as the pivot,

(M)(g)(1/3)(L) = (m2)(g)(L/2 - L/3)

M = (m2)/2

M = (115 kg)/2

M = 57.5 kg

57.5 - 43 kg = 14.5 kg

part 4 )

Mb*g/2 = 563.5 N

part 5 )

mg = 421.4 N

mg +Mb/2* g = 984.9 N

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