Rotational Statics questions. A gymnast with mass m 1 = 44 kg is on a balance be

Question

Rotational Statics questions.

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 = 108 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?

Explanation / Answer

Refer below figure,

1. Applying Newton’s second law we get,

F1+F2-Mg-mg =0 ---------------------(1)

Applying law of conservation at F2

m1g*(L-1/3) +F1*0 - m2g*(L/2-1/3) + F2*(L-2/3) =0 ---------------------(2)

44*9.8*(5-1/3) - 108*9.8*(5/2-1/3) +F2(5-2/3) =0        => F2= 64.8 N

Plugging this value in (1)

2. F1+64.8 -44*9.8-108*9.8 =0        => F1= 1424.8 N

3. For tipping beam we need F2=0N

Plug F=0 in (2)

(m1+m)g*(L-1/3) +F1*0 - m2g*(L/2-1/3) + F2*(L-2/3) =0

m is amount of extra mass

(44+m)*9.8*(5-1/3) - 108*9.8*(5/2-1/3) +F2*0 =0        => m = 6.14 N

4. Refer below figure,

F1*0 - m2g*(L/2-1/3) + F2*(L-2/3) - m1g*(L-2/3) +=0 ---------------------(2)

- 108*9.8*(5/2-1/3) +F2(5-2/3) - 44*9.8*(5-2/3) =0        => F2= 960.4 N

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