Question

A force of 25 N is applied at the end of a spanner whose effective length is 19.5 cm. Calculate the torque.​

Answer 1

$\fbox{\texttt{\green{Answer:}}}$

Torque = 4.875 Newton metres.

$\fbox{\texttt{\pink{Given:}}}$

• Force applied = 25 Newtons.

• Radius i.e., effective length = 19.5 cm = 0.195 metre.

$\fbox{\texttt{\blue{To\:find:}}}$

Torque(tendency of the force to turn or twist).

$\fbox{\texttt{\red{How\:to\:Find:}}}$

We know,

☛ Torque = Force × Radius

☛ Torque = 25 Newtons × 0.195 metre

☛ Torque = 25 × 0.195 Newton metres

☛ Torque = 4.875 Newton metres.

∴ Torque applied on the spanner is 4.875 N•m

Answer 2

Answer:

The Torque (τ) applied on the spanner is 4.875 m

Given:

1. Force Applied (F) = 25 N.

2. Length of spanner (r) = 19.5 cm

Explanation:

_____________________

Firstly converting Length of spanner in to m (S.I unit)

⇒ 1 cm = 10⁻² m

⇒ 19.5 cm = 19.5 × 10⁻² m

⇒ 19.5 cm = 0.195 m

⇒ r = 0.195 m

∴ We got the length of spanner in S.I units.

_____________________

_____________________

Torque: It is the cross product (or) Vector product of the Distance and Force.

From the formula we know,

⇒ τ = F × r

⇒ τ = 25 × 0.195

⇒ τ = 4.875

⇒ τ = 4.875 N-m

∴ The Torque (τ) applied on the spanner is 4.875 m

Note:

• Symbols have their usual meanings.

_____________________