Question

1. Find the moment of force of 20 N about an axis of rotation at a distance of 0.5m from the force​

Answer 1

TOPIC :- TORQUE

CONCEPT :- Torque about a point is defined as the cross product of radius vector from the point to the point of application of force and the force . Torque = r × F = Fr Sin∆ .Torque is known as moment of force.

STEPS :-

Torque = r × F

= 0.5 × 20

= 10 Nm Ans.

Answer 2

Question-

Find the moment of force of 20N about axis of rotation at distance 0.5m from the force.

Given-

• Force = 20N
• ⊥ Distance = 0.5m

To find-

• Moment of force

Calculation-

We know,

${\underline {\underline {\boxed {\red {\sf {Moment \: of \: force = Force \times ⊥ Distance }}}}}}$

$\sf {☞ \: Moment \:of \:force = F × ⊥ Distance}$

$\sf {☞ \: Moment \: of\: force = 20N × 0.5 m}$

$\sf { ☞ \:Moment \: of \:force = 10 Nm }$

Therefore, the moment of force of 20N about axis of rotation at distance 0.5m from the force is 10Nm.

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