Two blocks A and B,each of mass m,are placed | vedclass

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(D) Let the acceleration of the system be $a$. The net force acting on the system is $2F - F = F$ towards the left.Total mass of the system is $m + m

Vedclass · Accepted Answer

$3F$

Vedclass · Answer

(D) Let the acceleration of the system be $a$. The net force acting on the system is $2F - F = F$ towards the left.Total mass of the system is $m + m = 2m$.Therefore,the acceleration of the system is $a = \frac{F}{2m}$ towards the left.Now,consider the free body diagram of block $A$. The forces acting on $A$ are the applied force $F$ to the right,and the normal reaction $N$ from block $B$ acting perpendicular to the contact surface.The contact surface makes an angle of $30^{\circ}$ with the horizontal,so the normal $N$ makes an angle of $60^{\circ}$ with the horizontal.The component of $N$ in the horizontal direction is $N \cos 60^{\circ}$.Applying Newton's second law for block $A$ in the horizontal direction:$N \cos 60^{\circ} - F = ma$$N \left(\frac{1}{2}\right) - F = m \left(\frac{F}{2m}\right)$$N \left(\frac{1}{2}\right) - F = \frac{F}{2}$$N \left(\frac{1}{2}\right) = F + \frac{F}{2} = \frac{3F}{2}$$N = 3F$

Two blocks $A$ and $B$,each of mass $m$,are placed on a smooth horizontal surface. Two horizontal forces $F$ and $2F$ are applied on the two blocks $A$ and $B$ respectively,as shown in the figure. The block $A$ does not slide on block $B$. Then,the normal reaction acting between the two blocks is

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