A1.10-kg wrench is acting on a nut trying to turn it. the length of the wrench lies directly to the east of the nut. a force 150.0 n acts on the wrench at the position 15.0 cm from the center of the nut in a direction 30.0 degree north of east. what is the magnitude of the torque about the center of the nut ?

The problem can be solve using the formula

τ=r F sin(θ)

Where

τ is the torque,

r is the distance from where the pivot is to where the force is,

F is the force exerted, and

θ is the angle between the force and radius

τ=(0.15m) (150N) sin(30) = 11.25 Nm

gravity

explanation:

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