Acrowbar 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?

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

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,

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

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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0.863333

86%

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