Imagine a situation in which the horizontal s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the acceleration of the system be $a = F / (M + m + M_0)$.For block $m$ to be stationary with respect to $M_0$,the normal force $N$ exerted by

Vedclass · Accepted Answer

If $F = 0$,the blocks cannot remain stationary.

Vedclass · Answer

(A) Let the acceleration of the system be $a = F / (M + m + M_0)$.For block $m$ to be stationary with respect to $M_0$,the normal force $N$ exerted by $M_0$ on $m$ must provide the acceleration $a$. Thus,$N = ma$.The vertical forces on $m$ are gravity $mg$ downwards and friction $f$ upwards. For equilibrium,$f = mg$.Since $f \le \mu N$,we have $mg \le \mu (ma)$,which implies $a \ge g / \mu$.If $F = 0$,then $a = 0$,so $f = mg$ cannot be balanced,meaning the blocks cannot remain stationary. Thus,$(A)$ is correct.For $f = 0$,we need $mg = 0$,which is impossible. Thus,$(C)$ is incorrect.For a range of $F$ such that $a \ge g / \mu$,the blocks can remain stationary,not just for one unique value. Thus,$(B)$ is incorrect.Therefore,only statement $(A)$ is correct.

Imagine a situation in which the horizontal surface of block $M_0$ is smooth and its vertical surface is rough with a coefficient of friction $\mu$. Identify the correct statement$(s)$.

Similar Questions

$A$ block of mass $m$ is placed on the top of another block of mass $M$ as shown in the figure. The coefficient of friction between them is $\mu$. What is the maximum acceleration with which the block $M$ may move so that $m$ also moves along with it?

The maximum value of $F$ which can be applied on the system shown in the figure so that both blocks move together with the same acceleration is ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module