Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Solve the following :

Question:

Find the acceleration of the block of mass $M$ in the situation of figure. The coefficient of friction between the two blocks is $\mu 1$, and that between the bigger block and the ground is $\mu 2$.

Solution:

When a block M, moves with acceleration a towards right block $\mathrm{m}$ moves downwards and rightwards with acceleration $2 a$ and a respectively.

$N^{\prime}=M g+\mu_{1} N+T$ (Vertical) -(i)

$(T+T)-N-\mu_{2} N^{\prime}=M a$ (Horizontal) -(ii)

FBD of mass $\mathrm{m}$

N=ma (Horizontal) -(iii)

$m g-T-\mu_{1} N=m(2 a)$ (vertical) -(iv)

From (iii) and (iv)

$T=m g-2 m a-\mu_{1} m a-(\mathrm{v})$

Solving equation (i), (ii), (iii) and (iv)

$a=\frac{\left[2 m-\mu_{2}(M+m)\right] g}{M+m\left[5+2\left(\mu_{1}-\mu_{2}\right)\right]}$

 

 

Leave a comment

None
Free Study Material