If the hanging mass in problem 3 is 26 kg, what is the minimum value of L (to th

Question

If the hanging mass in problem 3 is 26 kg, what is the minimum value of L (to the nearest hundredth of a meter) for which the arrangement will be stable?

In problem 3, if L = 1.52 meters, what is the maximum value (to the nearest gram) of the hanging mass, m, for which the arrangement will be stable?

If the hanging mass in problem 3 is 26 kg, what is the minimum value of L (to the nearest hundredth of a meter) for which the arrangement will be stable? In problem 3, if L = 1.52 meters, what is the maximum value (to the nearest gram) of the hanging mass, m, for which the arrangement will be stable?

Explanation / Answer

Part A)

At the minimum stability, F1 will go to zero. F2 will be the pivot point and we can solve with the sum of torques

(Mg)(L-1) = mg(2-L) g cancels

100L - 100 = 52 - 26L

126L = 152

L = 1.21 m

Part B)

Same idea here

100(1.52-1) = m(2-1.52)

52 = .48m

m = 108.333 kg (That is 108333 grams)

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