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Leona [35]
3 years ago
12

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

object at the short​ end?
Physics
1 answer:
NikAS [45]3 years ago
4 0

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

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