A gymnast with mass m1 = 42 kg is on a balance beam that sits on (but is not att

Question

A gymnast with mass m1 = 42 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 106 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?

THANKS~

Explanation / Answer

forces upward = forces downward (S1) + (S2) = (m1)(g) + (m2)(g) (S1) + (S2) = (42 kg)(9.81 m/s²) + (106 kg)(9.81 m/s²) (S1) + (S2) = 412.02+1039.86 (S1) + (S2) = 1452 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[(42 kg)+ (106 kg)/2](9.81 m/s²) (S1)(2) + (S2) = 2796 N [2] 1)What is the force the left support exerts on the beam? [2] - [1]: (S1) = 2796 - 1452 N (S1) = 1344 N 2)What is the force the right support exerts on the beam? 1344 N + (S2) = 1452 N (S2) = 108 N 3)How much extra mass could the gymnast hold before the beam begins to tip? Using the left support as the pivot, (M)(g)(1/3)(L) = (m2)(g)(L/2 - L/3) M = (m2)/2 M = (106 kg)/2 M = 53 kg 53 - 42 kg = 11 kg Feel free to revert for any further clarification Cheers PK!

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