Question

Find the magnitude of the torque produced by a 4.5 N force applied to a door at a perpendicular distance of 0.26 m from the hinge. Answer in units of N · m.

Answer 1

Answer:

The required torque is 1.17 N-m.

Explanation:

The given data :-

The magnitude of force ( F ) = 4.5 N.

The length of arm ( r ) = 0.26 m.

Here given that force is applied at perpendicular means ( ∅ ) = 90°.

The torque ( T ) is given by

T = F * r * sin∅

T = 4.5 * 0.26 * sin 90°

T = 4.5 * 0.26 * 1

T = 1.17 N-m.

Answer 2

The magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.

Given the data in the question;

• Force; $F = 4.5N$

• Perpendicular distance or radius; $r = 0.26m$

• Since the force is perpendicular to the lever, Angle; $\theta = 90^o$

Torque; $T = \ ?$

Torque simply the measure of the force that can cause an object to rotate about an axis. It is expressed as:

$T = rFsin\theta$

Where r is radius, F is force applied and θ is the angle between the force and the lever arm.

We substitute our given values into the equation;

$T = 0.26m * 4.5N * sin90^o\\\\T = 0.26m * 4.5N * 1\\\\T = 1.17N.m$

Therefore, the magnitude of the torque produced by the given force perpendicular to the lever is 1.17N.m.

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