Question

A cart of mass M has a block of mass m attached to it as shown in fig. The coefficient of friction between the block and the cart is μ. What is the minimum acceleration of the cart so that the block m does not fall?(a) μg(b) g/μ(c) μ/g(d) Mμg/m

Answer 1

component of force F horizontally forward = F Cos B

vertical component of F, upwards = F sin B

Normal reaction N of the floor on the block = N = m g - F Sin B,  as N+ F Sin B is balanced by N.  There is no movement of the block in the vertical direction.

kinetic friction on the block = μ N = μ (m g - F Sin B)  acting backwards.

net force on the block M = F Cos B - μ (m g - F Sin B)

acceleration of hte block in the forward direction :

           force / mass = [ F Cos B - μ (m g - F Sin B) ] /M

Answer 2

Answer:

B) g/μ

Explanation:

Mass of the cart = M (Given)

Coefficient of friction between the block and the cart = μ. (Given)

If the cart acceleration which is = ′α′

Then the, Pseudo force on mass F = mα

Force of friction will be = μN

For horizontal equilibrium, it will be F=N

Thus, f = μmα

The block will not fall as long as f ≥ mg

= μmα ≥ mg

= α ≥ g/μ

Thus, the minimum acceleration of the cart so that the block m does not fall is g/μ