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In opening a door, a 35.8 N force is applied. Find the torque produced by this force with respect to point O. (Let [tex]r = 0.920 \, \text{m}[/tex] and [tex]\theta = \text{given angle}[/tex].)

Answer :

Final answer:

The torque (τ) produced by a 35.8 N force applied perpendicularly to a door at a distance (r) of 0.920 m from the pivot point is calculated using the formula τ = rF which gives a value of τ = 32.936 N·m.

Explanation:

The question requires us to calculate the torque produced by a force acting on a door with reference to a specified point. In the field of physics, torque is calculated using the formula τ = rFsinθ, where τ is the torque, r is the distance from the pivot point to where the force is applied, F is the force, and θ is the angle between the force and the vector directed from the point of application to the pivot point.

In the given scenario, the force (F) is 35.8 N, the radius or distance (r) is 0.920 m, and the force is applied perpendicularly, which implies that θ is 90 degrees.

When θ is 90 degrees, sinθ is 1. Thus, we can simplify the formula for torque to τ = rF. Substituting in the given values, we get τ = (0.920 m)(35.8 N) = 32.936 N·m.

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