Question

Four identical $2kg$ blocks are arranged as shown in the figure. All the surfaces are rough.

A force of $10N$ is applied to block C parallel to incline as shown. Rank the normal force on the bottom surface of each block?

• A. $N_D<N_A<N_C<N_B$
• B. $N_D>N_A>N_C>N_B$
• C. $N_D<N_C<N_B<N_A$
• D. $N_D<N_B<N_C<N_A$

Answer 1

The correct option is A $N_D<N_A<N_C<N_B$

Let the normal force on the bottom surface of blocks A, B, C and D be $N_A,N_B,N_C$ and $N_D$ respectively.

Given:

• m = 2 kg
• F = 10 N

A	B	C	D
$N_A=mg$	$N_B=2mg$	$N_C=mg+F\sin \theta$	$N_D=mg\cos \theta$
$N_A=2g$	$N_B=4g$	$N_C=2g+10\sin \theta$	$N_D=2g\cos \theta$

1. Comparing $N_A$ and $N_B$
   $2g<4g\Rightarrow N_A<N_B$
2. Comparing $N_C$ and $N_D$
   If, $\theta =0,N_C=N_D$
   $\theta =\frac{\pi }{2},N_C>N_D$
   Since cosine is decreasing function and sine is increasing function, $N_C>N_D$ in general.
3. Comparing $N_B$ and $N_C$
   Maximum value of sine function is 1 at $\theta =\frac{\pi }{2}\Rightarrow 4g>2g+10$
   Hence in general $N_B>N_C$
   
4. Comparing $N_A$ and $N_D$
   As cosine function is decreasing function, in general $N_A>N_D$

Combining these results $N_D<N_A<N_C<N_B$

Hence the right option is A.