A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. what is the torque on the bar?

The torque on the bar is  - 9.375 N

From the information given:

The diagrammatic expression of the question given can be seen in the image attached below, from the image the direction of the torque on the bar is in a clockwise direction as such, we will take the negative sign into consideration.

∴

The torque can be computed by using the formula:

T = - (F × d × sin θ)

T = - (25 N × 0.75 m × sin 30)

T = - (25 N × 0.75 m × 0.5)

T = -9.375 N

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Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

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