A wooden wedge of mass M and inclination angl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the acceleration of the wedge be $a$ towards the left. Since the block is stationary with respect to the wedge,the block also moves with the s

Vedclass · Accepted Answer

$(M+m) g 	an \alpha$

Vedclass · Answer

(A) Let the acceleration of the wedge be $a$ towards the left. Since the block is stationary with respect to the wedge,the block also moves with the same acceleration $a$ towards the left.For the entire system (wedge + block),the total force is $F = (M+m) a \dots (i)$.Now,consider the free body diagram of the block of mass $m$ in the non-inertial frame of the wedge. The forces acting on the block are:$1$. Gravitational force $mg$ downwards.$2$. Normal force $N$ perpendicular to the inclined surface.$3$. Pseudo force $ma$ acting horizontally towards the right.For the block to remain stationary on the wedge,the component of the pseudo force along the incline must balance the component of the gravitational force along the incline:$ma \cos \alpha = mg \sin \alpha$$a = g \frac{\sin \alpha}{\cos \alpha} = g 	an \alpha \dots (ii)$Substituting equation $(ii)$ into equation $(i)$,we get:$F = (M+m) g 	an \alpha$.

$A$ wooden wedge of mass $M$ and inclination angle $\alpha$ rests on a smooth floor. $A$ block of mass $m$ is kept on the wedge. $A$ force $F$ is applied on the wedge as shown in the figure such that the block remains stationary with respect to the wedge. The magnitude of the force $F$ is:

Similar Questions

$A$ block of mass $m$ is resting on a wedge of angle $\theta$ as shown in the figure. The wedge is given a horizontal acceleration $a$. What is the minimum value of $a$ so that the mass $m$ falls freely?

$A$ $60\, kg$ man stands on a spring scale in a lift. At some instant,he finds the scale reading has changed from $60\, kg$ to $50\, kg$ for a while and then comes back to the original mark. What should we conclude?

$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration $a$. Then:

Find the force $F$ required on the cart so that the $2 \ kg$ and $1 \ kg$ blocks remain stationary relative to the cart.

$A$ man of mass $60\, kg$ records his weight on a weighing machine placed inside a lift. The ratio of weights of the man recorded when the lift is ascending with a uniform speed of $2\, m/s$ to when it is descending with a uniform speed of $4\, m/s$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module