Question

Shiv opens a nut with the help of a spanner of length 20 cm by applying a force of 60 N. Calculate the moment of force required to open the nut.

Answer 1

Answer-

Length= 30 Cm , Force= 20 N

Moment of force= F×D

=30×20

=600NM

Answer 2

Answer:

• Moment of force required to open the nut is 12 Nm.

Explanation:

Given

Shiv opens a nut with a spanner of length 20 cm by applying force 60 N, so

• Length of spanner, s = 20 cm
• Force applied, F = 60 N

To find

• Moment of force required to open the nut = Torque, τ =?

Knowledge required

• Formula to calculate Torque (Torque means moment of force)

        τ = F s  sin θ

[ Where τ represent torque in, F is force and s is distance, θ is the angle difference between F and s ]

• Torque is a vector quantity.
• SI unit of torque : N m    [Newton metre]
• 1 N m = 1 kg m² s⁻¹
• Dimensional formula of Torque : [ M L² T⁻² ]

Calculation

Converting length of spanner given in cm into m

→ s = 20 cm

→ s = 20/100  m

→ s = 0.2 m

In this case of opening the nut , θ will be equal to 90°. so

Now, Using formula to calculate moment of force (Torque)

→ τ = F s  sin θ

→ τ = ( 60 N ) · ( 0.2 m ) ( sin 90° )

→ τ = ( 60 N ) ( 0.2 m ) ( 1 )

→ τ = ( 60 N ) ( 0.2 m )

→ τ = 12 N m

Therefore,

• Moment of force required to open the nut is 12 N m.