Question

A cart of mass $M$ moves with acceleration $a$ as shown in the figure. The minimum force required to hold a block of mass $m$ together is $($friction coefficient between two blocks is $\mu )$

• A. $\frac{(M-m)g}{\mu }$
• B. $\frac{(M+m)g}{\mu }$
• C. $\frac{Mg}{\mu }$
• D. $\frac{mg}{\mu }$

Answer 1

The correct option is B $\frac{(M+m)g}{\mu }$

Since it is given that the block is stationary w.r.t the cart, we can consider the cart frame of reference with a pseudo force $ma$ acting on the block. The FBD will be


Writing the force balance equations,

$f=mg$
$N=ma$

Since we want the minimum acceleration of the cart for which block will remain at rest, we have to consider the limiting value of friction which can balance the weight of the block.

Thus,
$f=\mu N=\mu ma=mg$

$a=\frac{g}{\mu }$

Therefore, minimum force required to hold the block stationary is
$F_{min}=(M+m)\frac{g}{\mu }$