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

Learn more about torque here:

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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

no simpler subs form complex ones. synthesise = make

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!