Multiple Choice Questions (MCQs): Friction - JEE MCQ

Frictional forces act over the common surfaces of the two bodies to avoid and restrict relative moment between those two surfaces.

When the body is in rest it is under static friction but when it starts moving (neither rolling nor sliding), the static friction slowly chngs to kinetic friction as the coefficient of static friction start decreasing and that of kinetic friction starts increasing. In case it starts rolling motion then the friction is rolling friction & if it slides then sliding fiction.

The impending motion refers to the state of a body when it is on the verge of slipping. In such cases, the static friction has reached its upper limit and is given by the equation, F=μ_sN

Limiting friction, f = μmg = 0.6 × 1 × 9.8 = 5.88 N

Applied force = F = ma = 1 × 5 = 5 N

As F < f, so force of friction = 5 N

We know that centripetal acceleration, c = v²/r
Thus we get r = v²/c
= 36 / 120
= .3 /m

At the time when the block just starts to move, we get that net force acting upon it is 0, thus we get, f - mg.sin a = 0
Where f is friction force and a is angle of incline.
We also know that f = 1√3 x N
= 1√3 x mg.cos a
Thus we get 1√3 x mg.cos a = mg.sin a
Thus we get tan a = 1√3
And a = 30

Mass of the block = 2 kg

Weight of the block = mg = 2 × 9.8 = 19.6 N

The component of the weight along normal = 19.6 Cos 30° = 16.97 N

Hence, the normal force (N) = 16. 97 N

Now, the component of the weight along the inclined plane = 19.6 Sin 30°

Along the inclined plane = 9.8 N

Now, the friction = μ N = 0.7 × 16.97 = 11.87 N

Since, 9.8 N is less than the maximum value of the friction (11.87 N)

Hence, the frictional force = 9.8 N

We know that centripetal acceleration, a = v x v /r
Here v = 6m/s and r = 0.3m
Thus a = 36 / 0.3 m/s²
= 120 m/s²

Frictional force is the opposing force which plays between two surfaces and it destroys the relative motion between them. Frictional force is a non-conservative force. The force produced by two surfaces that contact and slide against each other, that force is called the frictional force. These forces are affected by the nature of the surface and amount of force acting on them.

In case of a bicycle, the front wheel of the bicycle is connected to a rod passing through its centre. The force acting on the wheel about its central axis by the force coming from the rest of the bicycle is zero. Front wheel obtains linear velocity by pedalling but it cannot rotate it.

Wheel or ball can also be rolled by pushing on it. The frictional force prevents the wheel from sliding forward at the point of contact. Here, the frictional force prevents the wheel from sliding forward and it is in the opposite direction.

So, in the case of the wheel, the point P which is in contact with the ground tries to go backward due to rotation. Frictional force will oppose this motion. Hence it will move forward.
Hence the direction of frictional force at the point P of the wheel is in forward direction.
Note: Frictional force opposes the motion. Here static friction holds a wheel or a ball on the surface. Frictional force is equal and opposite in direction to the applied force parallel to the contacting surfaces. The resistance due to the rolling body on a surface is called rolling friction. Torque is a force that acts on a body that is undergoing rotation.