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- Two smooth blocks are placed at a smooth corner as shown in figure. Both the blocks are having mass m. We apply a force F on the small block m. Block A presses block B in the normal direction, due to which pressing force on vertical wall will increase, and pressing force on the horizontal wall decreases, as we increases F(θ=37°. with horizontal). <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/fd417b1e0bb30121932d4d3c8d0847ae-.png /> As soon as the pressing force on the horizontal wall by block B becomes zero, it will lose contact with ground. If the value of F further increases, block B will accelerate in the upward direction and simultaneously block A will move towards right. If both the blocks are stationary, the force exerted by ground on block A is
Q.
Two smooth blocks are placed at a smooth corner as shown in figure. Both the blocks are having mass $m$. We apply a force $F$ on the small block $m$. Block $A$ presses block $B$ in the normal direction, due to which pressing force on vertical wall will increase, and pressing force on the horizontal wall decreases, as we increases $F\left(\theta=37^{\circ}\right.$ with horizontal).
As soon as the pressing force on the horizontal wall by block $B$ becomes zero, it will lose contact with ground. If the value of $F$ further increases, block $B$ will accelerate in the upward direction and simultaneously block $A$ will move towards right.
If both the blocks are stationary, the force exerted by ground on block $A$ is
Laws of Motion
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Solution:
If both the blocks are stationary.
Balancing forces along $x$-direction
$F =N \sin \theta$
$\Rightarrow N =F / \sin \theta$
Balancing forces along $y$-direction.
$N_{y} =m g+ N \cos \theta$
$=m g+\left(\frac{F}{\sin \theta}\right) \cos \theta=m g+F \cot \theta$
$N_{y} =m g+\frac{4 F}{3}$
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