Question

A body of mass $10\text{kg}$ is placed on a smooth inclined plane as shown in the figure. The inclined plane is moved with a horizontal acceleration $a$.
The normal reaction between block and inclined plane is

• A. $92\text{N}$
• B. $44\text{N}$
• C. $56\text{N}$
• D. Can't be determined

Answer 1

The correct option is C $56\text{N}$

FBD of block w.r.t. inclined plane
$\sum F_x=0\Rightarrow T\cos {37}^{\circ }-N\sin {37}^{\circ }-ma=0$
$\Rightarrow 4T-3N=200$ ... (i)
$\sum F_y=0\Rightarrow N\cos {37}^{\circ }+T\sin {37}^{\circ }-10g=0$
$\Rightarrow 4N+3T=500$ ... (ii)
By solving equation (i) & (ii), $N=56$ N