Question

Calculate the moment of a force of 16N about an axis of rotation at a normal distance of 0.5m from the force?

Answer 1

Given : Force = 16N

Foce of 16N rotate about an axis of rotation at a normal distance 0.5 m. Means,

Perpendicular distance = 0.5 m

Find : We have to find the moment of force. Means, Torque.

Solution :

Torque = Force × perpendicular distance

Torque is defined as product of force and the perpendicular distance of the force from the axis of rotatio.

i.e.

$\tau \: = \: F\: \times \: \perp \: dist$

or

$\tau \: = \: F\: \times \: s$

Put the known values in above formula

$\Rightarrow\:\tau\:=\:16\:\times\:0.5$

$\Rightarrow\:\tau\:=\:8.0$

$\Rightarrow\:\tau\:=\:8$

Unit of torque :

Unit of Force = N (newton)

Unit of distance = m (meter)

So,

Unit of torque = N × m

$\Rightarrow$ N-m

Moment of force is 8 N-m.

Answer 2

$\huge\underline\mathfrak\blue{Answer}$

force = 16N

distance = 0.5m

Moment of force = 16 × 0.5 Nm

Therefore

Moment of force = 8 Nm

Thanks!! ✨✨❤