Question

A wrench is used to tighten a nut. A 15N perpendicular force is applied 50cm away from the axis of rotation, and moves a distance of 10 cm as it turns. What is the torque applied to the wrench?

Answer 1

Answer:

• Torque applied to the wrench = 7.5 Nm

Explanation:

Given,

• Perpendicular force applied, F = 15 N
• Distance from the axis of rotation, r = 50 cm = 0.5 m
• Distance moved by wrench when turns = 10 cm

We need to find,

• Torque applied, τ =?

Formula to calculate torque is given by,

      τ = F r sin θ

[ Where, τ is torque, F is force, r is distance from the axis of rotation, θ is the angle between F and r ]

For the given condition,

Since, Force applied by the wrench is perpendicular to the distance from axis of rotation therefore, θ = 90°

Using the formula for torque

→ τ = F r sin θ

→ τ = 15 × 0.5 × sin 90°

→ τ = 15 × 0.5 × 1

→ τ = 7.5 Nm

Therefore,

Torque applied to the Wrench will be 7.5 Nm.

Note: Torque (τ) depends on three factors: magnitude of force applied (F), direction of force (θ) and position of application of force (r).

Answer 2

Answer:

Given :-

• A wrench is used to tighten a nut. A 15 N perpendicular force is applied is 50 cm away from the axis of rotation, and moves a distance of 10 cm as it turns.

To Find :-

• What is the torque applied to the wrench.

Formula Used :-

To find torque, we know that,

$\sf\boxed{\bold{\large{\tau =\: rF\: sin{\theta}}}}$

where,

• τ = Torque
• r = Radius
• F = Force
• $\theta$ = Angle between F and r

Solution :-

Given :

• Force (F) = 15 N
• Radius (r) = 50 cm = $\sf\dfrac{50}{100}$ = 0.5 m
• $\theta$ = 90°

According to the question by using the formula we get,

⇒ $\sf \tau =\: 0.5 \times 15 \times sin\: {90}^{\circ}$

(As we know that, sin 90° = 1)

⇒ $\sf \tau =\: 0.5 \times 15 \times 1$

⇒ $\sf \tau =\: 7.5 \times 1$

➠ $\sf\bold{\red{\tau =\: 7.5\: Nm}}$

$\therefore$ The torque applied to the wrench is 7.5 Nm .