Question

A cart of mass \[M\] has a block of mass \[m\] in contact with it as shown in the figure. The coefficient of friction between the block and the cart is \[\mu \] . What should be the minimum acceleration of the cart so that the block of mass \[m\] does not fall?

Answer 1

Hint: In this question, let us assume the cart of mass \[M\] moves with an acceleration \[a{\text{ m/}}{{\text{s}}^2}\]. Thus, a pseudo force \[ma\] acts on the small block of mass \[m\] and it must be equal to the normal force between the blocks for horizontal equilibrium. The force of friction \[\mu N\] will act in the upward direction and must be equal to \[mg\] so that the block does not fall. From this condition, we can find out the minimum acceleration required.

Formula used:
\[F = ma\]
Where \[F\] is the total force, \[m\] is the mass of the system and \[a\] is the net acceleration of the system.
\[f = \mu N\]
Where, \[f\] is the limiting frictional force, \[\mu \] is the coefficient of friction and \[N\] is the normal force acting.

Complete step by step answer:
Let us assume the cart of mass \[M\] is moving with an acceleration of \[a{\text{ m/}}{{\text{s}}^2}\]. It is in contact with a block of mass \[m\] such that the coefficient of friction between the cart and block is \[\mu \]. A normal force \[N\] acts on the block in the horizontal direction.As the cart is accelerating a pseudo force \[F = ma\] acts in the opposite direction of acceleration
For horizontal equilibrium,
\[N = F = ma\]

Frictional force \[f = \mu N\] acts in the upward direction. For block to be in equilibrium frictional force must balance its weight,
\[f \geqslant mg\]
\[ \Rightarrow \mu N \geqslant mg\]
Substituting \[N = F = ma\],
\[ \Rightarrow \mu ma \geqslant mg\]
\[ \therefore a \geqslant \dfrac{g}{\mu }\]

Thus, minimum acceleration so that the block does not fall must be \[\dfrac{g}{\mu }{\text{ m/}}{{\text{s}}^2}\].

Note: Pseudo force is an imaginary force like frictional force and comes in effect when the frame of reference has started acceleration compared to a non-accelerating frame. It always acts in the opposite direction of the motion. The work done by a pseudo force is zero as it acts to appear on the body.