Question

There is no friction anywhere in the system shown in figure. The pulley is light. The wedge is free to move on the frictionless surface. A horizontal force $F$ is applied on the system in such a way that $m$ does not slide on $M$ or both move together with some common acceleration. Given $M>\sqrt{2}m$.

Match the entries of Column I with that of Column II.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{i.Pseudo-force acting on m as seen from the frame of M is}& \text{a. equal to}\frac{mF}{m+M}\\ \text{ii.Pseudo-force acting on M as seen from the frame of m is}& \text{b. greater than}\frac{mF}{m+M}\\ \text{iii. Normal force (for}\theta ={45}^{\circ })\text{between m and M is}& \text{c. less than}mg\sin \theta \\ \text{iv. Normal force between ground and M is}& \text{d. greater than}mg\sin \theta \end{array}$

• A. $\text{i- a, c; ii- b, c; iii- b, c; iv- b, d}$
• B. $\text{i- b, c; ii- a, c; iii- b, c; iv- b, d}$
• C. $\text{i- b, c; ii- a, c; iii- b, d; iv- b, c}$
• D. $\text{i- b, c; ii- b, d; iii- a, c; iv- b, c}$

Answer 1

The correct option is A $\text{i- a, c; ii- b, c; iii- b, c; iv- b, d}$

As the blocks move together with same acceleration:
$a=\frac{F}{M+m}$

Let $R$ be the reaction force on wedge by the ground.
Here, $R=(M+m)g$

The FBD of the wedge:

The FBD of the block:

For (i):
Pseudo force acting on $m$:
$F_i=\frac{mF}{m+M}$
Hence (a) matches.

We can also say that,
$F=ma\cos \theta +mg\sin \theta$
$F=F_i\cos \theta +mg\sin \theta$
$\frac{F_i(M+m)}{m}=F_i\cos \theta +mg\sin \theta$
By solving this, I can say that
$F_i<mg\sin \theta$
Hence (c) matches.

For (ii):
Pseudo force acting on $M$:
$F_{ii}=Ma$
But, $Ma>ma$, so $F_{ii}>F_i$
$F_{ii}>\frac{mF}{m+M}$
$F_{ii}=\frac{MF}{m+M}$
Hence (b) matches.

Moreover, lets take
$F=ma\cos \theta +mg\sin \theta$
$F=mMa\cos \theta /M+mg\sin \theta$
$\frac{F_{ii}(M+m)}{M}=\frac{F_{ii}m\cos \theta }{M}+mg\sin \theta$
By solving this, we can say that
$F_{ii}<mg\sin \theta$
Hence (c) matches.

For (iii): $(\theta ={45}^{\circ })$
If we balance forces in FBD of $M$, we get
$N\sin \theta =Ma$
$N/\sqrt{2}=Ma$
$N=\sqrt{2}Ma$
$N=\sqrt{2}{F}_{ii}$
As, $F_{ii}>\frac{mF}{m+M}$
$N>\frac{mF}{m+M}$
Hence (b) matches.

Again, lets take previously derived equation
$\frac{F_{ii}(M+m)}{M}=\frac{F_{ii}m\cos \theta }{M}+mg\sin \theta$
But, $F_{ii}=N/\sqrt{2}$
Hence, we get
$\frac{N(M+m)}{\sqrt{2}M}=\frac{Nm\cos \theta }{\sqrt{2}M}+mg\sin \theta$
By solving this, we can say that
$N<mg\sin \theta$

For (iv):
We have $R=(M+m)g$
$R>mg$
$R>mg\sin \theta$
Hence (d) matches.

Moreover, we know that $F_i<mg\sin \theta$
Hence, $R>mg\sin \theta >F_i$
$R>\frac{mF}{m+M}$
Hence (b) matches.

So combining all answers we can say option 'a' is correct.
$\text{i- a, c; ii- b, c; iii- b, c; iv- b, d}$