A gymnast with mass m1 = 46 kg is on a balance beam that sits on (but is not att

Question

A gymnast with mass m1 = 46 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 117 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?

Explanation / Answer

Here ,

m1 = 46 Kg

m2 = 117 Kg

Length , L = 5 m

1) let the force applied by the left support is Fl

balancing the moment of forces about the right suppoert

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

Fl/3 - 117 * 9.8/6 - 46 * 9.8 * 2/3 = 0

solving for Fl

Fl = 1474.9 N

the force on the left support is 1474.9 N

b)

balancing the vertical force ,

Fl + Fr = m1 * g + m2 * g

1474.8 + Fr = 117 * 9.8 + 46 * 9.8

Fr = 122.6 N

the force on the right support is 122.6 N

3)

let the extra mass that he can hold is m

for just tipping , the support from the right beam = 0

balancing the moment of forces about the left support

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

(46 + m )/3 = 117/6

m = 12.5 Kg

he can hold 12.5 Kg extra mass

4)

balancing the force about right support

Fl * L/3 - 117 * 9.8 * (1/6) = 0

Fl = 573.3 N

the force on left support is 573.3 N

5)

for the force on the right support is Fr

balancing the moment of forces about the left support

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

-46 * 9.8/3 - 117 * 9.8/6 + Fr/3 = 0

Fr = 1024.1 N

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