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 = 116 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) 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?
4) What is the force the right support exerts on the beam?

Explanation / Answer

1) Refer below figure,

Applying law of conservation of torque at right support,

F1*0 – m1g*0 +F2*d1-m2g*d2 = 0

F2*d1 – m2g*d2 = 0 ------------(1)

Applying Newton’s second law along vertical

F1+F2 –m1g-m2g=0 ---------(2)

d1=(5/2)-(1/3) = 2.17m

d2= 5-2*1/3 = 4.33m

F2*4.33– 116*9.8*2.17 = 0    => F2=569.7 N

From (2)

569.7 + F1 – 46*9.8 – 116*9.8 = 0     => F1= 1018 N

2) Refer below figure,

Applying law of conservation of torque at right support,

F1*d1 – m2g*d2 +F2*0+m1g*0 = 0

F1*d1 – m2g*d2 = 0 ------------(1)

Applying Newton’s second law along vertical

F1+F2 –m1g-m2g=0 ---------(2)

d1=(5/2)-(1/3) = 2.17m

d2= 5-2*1/3 = 4.33m

F1*4.33– 116*9.8*2.17 = 0    => F1=569.7 N

From (2)

569.7 + F2 – 46*9.8 – 116*9.8 = 0     => F2= 1018 N

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