Question

The moment of force of 20 N about a point is 4 Nm . Calculate the distance of a point of application of force fro the point.

Answer 1

Explanation:

Given : F=5 N

Moment of force (or torque ) τ=20 Nm

Let the distance of point of application of force from point P be d.

∴ τ=Fd

OR 20=5×d ⟹d=4 m

Answer 2

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✤ Required Answer:

✒ GiveN:

• Moment of force = 4 Nm

• Force on the body = 20 N

✒ To FinD:

• Distance of point of application of force from the point....??

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✤ How to solve?

For this, Let's know what is moment of force /or torque.

The moment of force brings the turning effect on a body about an axis. It is the product of magnitude of force and the perpendicular distance of line of action of force about the axis of rotation.

So, It is given by :

Moment of force about point of axis of rotation O = Force × Perpendicular distance of force from point O

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✤ Solution:

We have,

• Moment of force = 4 Nm

• Force = 20 N

|| Let the distance of point of application of force from the point of axis O be d ||

By using formula,

➝ Moment of force = Force × Distance

➝ 4 Nm = 20 N × d

➝ d = 4 Nm/20 N

➝ d = 0.2 m

❍ So, Distance from point of application = 0.2 m

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