Question

In the figure given below, two blocks of mass $m$ and $M$ are connected through a light inextensible rope. Then, find the tension in the connecting rope. Consider coefficient of static friction between the inclined surface and the blocks to be $\mu$.

• A. $(M+m)g\sin \theta$
• B. $(M+m)g\sin \theta -\mu mg\cos \theta$
• C. $\text{Zero}$
• D. $(M+m)g\cos \theta$

Answer 1

The correct option is C $\text{Zero}$

From FBD:


Applying equilibrium condition perpendicular to the inclined surface
$N_1=Mg\cos \theta$ &
$N_2=mg\cos \theta$

Let us assume that both blocks move with acceleration $a$ down the incline to maintain the string constraint.
For block $M$, equation of dynamics is:
$Ma=Mg\sin \theta -T-f_1$
where $f_1=\mu N_1=\mu Mg\cos \theta$ (kinetic friction)
$\Rightarrow Ma=Mg\sin \theta -T-\mu Mg\cos \theta$
$\therefore a=g\sin \theta -\frac{T}{M}-\mu g\cos \theta ....\left(i\right)$

For block $m$, equation of dynamics is:
$ma=mg\sin \theta +T-f_2$
where $f_2=\mu N_2=\mu mg\cos \theta$
$\Rightarrow ma=mg\sin \theta +T-\mu mg\cos \theta$
$\therefore a=g\sin \theta +\frac{T}{m}-\mu g\cos \theta ....\left(ii\right)$

Equating Eq. $\left(i\right)$ and $\left(ii\right)$:
$g\sin \theta -\frac{T}{M}-\mu g\cos \theta =g\sin \theta +\frac{T}{m}-\mu g\cos \theta$
$\Rightarrow \frac{T}{M}+\frac{T}{m}=0....\left(iii\right)$
Since mass can never be zero for the blocks, for validating Eq. $\left(iii\right)$
$T=0$
$\therefore$option $\left(C\right)$ is correct.