Question

A wheel of diameter 2m can be rotated about an axis passing through its centre by a moment of force equal to 2.0 N m. what minimum force must be applied on its rim?​

Answer 1

Answer:

The minimum force must be applied on its rim is 2N.

Explanation:

Given,

The diameter of a wheel = 2m

The moment of force = 2.0Nm

To find,

The minimum force.

Solution,

We know that the radius is equal to the half of the diameter. Therefore,

Radius, OP = $\frac{2}{2} = 1$m

Perpendicular distance, OP = 1m

We know that the moment of force is equal to the product of force and the perpendicular distance. i.e.

$\Rightarrow \rm{Moment \; of \; force = Force \times Perpendicular \; of \; distance}$

By substituting the known values in the formula, we get:

$\begin{array}{l}\Rightarrow 2.0 = F \times 1 \\ \Rightarrow F = \frac{2}{1} \\ \Rightarrow = F = 2\end{array}$

Hence, the minimum force must be applied on its rim is 2N.

Answer 2

Explanation:

Given that wheel of diameter 2m can be rotated about an axis passing through its centre by a moment of force equal to 2.0 N m.

We need to find out the minimum force applied on its rim. Let's say the minimum force applied on its rim is x.

Diameter of wheel = 2 m

Radius = Diameter/2 = 2m/2 = 1 m

Radius is equal to the distance. So, the distance is 1 m.

Now,

Momentum of force is defined as the product of force and perpendicular distance.

Momentum of force = Force × distance

Substitute the known values,

2 = x × 1

2/1 = x

x = 2

Therefore, the minimum force applied on its rim is 2N.