Question

A block of mass m is placed on an inclined plane as shown in figure. What must be the frictional force between block and incline if the block is not to slide along the incline when the incline is accelerating to the right at $3m/s^2(\sin {37}^{\circ }=\frac{3}{5})?(Takeg=10m/s^2)$

• A. $\frac{9}{25}mg$
• B. $\frac{28}{50}mg$
• C. $\frac{21}{50}mg$
• D. $\frac{31}{50}mg$

Answer 1

The correct option is A $\frac{9}{25}mg$

For block to remain stationary w.r.t. incline, it's relative acceleration must be zero.
Changing the frame to incline, since it is non inertial frame of reference we apply a pseudo force of magnitude

$\to$$m\times accelerationofnoninertialframe$ in opposite direction of incline acceleration.
From $FBD$
Balancing force along incline plane
$\to$$mgsin{37}^{\circ }=m\times 3cos{37}^{\circ }+f$...(i)
balancing forces perpendicular to incline plane
$\to$$mgcos{37}^{\circ }+m\times 3sin{37}^{\circ }=N$...(ii)
From equation (i) value of friction force:
$\therefore$ $f=\frac{9}{25}mg$