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You're carrying a 3.6-m long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. How much force must you exert to keep the pole motionless in a horizontal position?

Express your answer in newtons.

User Rvnovaes
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1 Answer

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Final answer:

To keep the pole motionless in a horizontal position, you need to exert a force equal to the weight of the pole as well as the torque created by the weight of the pole and its length. The force required is 22.87 N.

Step-by-step explanation:

To keep the pole motionless in a horizontal position, you need to exert a force equal to the weight of the pole as well as the torque created by the weight of the pole and its length.

The weight of the pole can be calculated using the formula

F = mg,

where F is the weight, m is the mass, and g is the acceleration due to gravity. In this case, the weight of the pole is

24 kg * 9.8 m/s² = 235.2 N.

The torque created by the weight of the pole can be calculated using the formula

Torque = force * distance.

The distance in this case is 35 cm = 0.35 m.

Therefore, the torque is

235.2 N * 0.35 m = 82.32 Nm.

To keep the pole motionless, you need to balance the torque created by the weight of the pole with an equal and opposite torque. This can be achieved by exerting a force at the end of the pole. The force required can be calculated using the formula Force = torque / distance. In this case, the distance is 3.6 m.

Therefore, the force required is 82.32 Nm / 3.6 m = 22.87 N.

User Davy Landman
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