A block of mass m lies on wedge of mass M, which lies on fixed horizontal surface. The wedge is free to move on the horizontal surface. A horizontal force of magnitude F is applied on block as shown, neglecting friction at all surfaces, the value of force F such that block has no relative motion w.e.t wedge will be: (where g is acceleration due to gravity)

Question

A block of mass m lies on wedge of mass M, which lies on fixed horizontal surface. The wedge is free to move on the horizontal surface. A horizontal force of magnitude F is applied on block as shown, neglecting friction at all surfaces, the value of force F such that block has no relative motion w.e.t wedge will be: (where g is acceleration due to gravity) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5d5b8f266cab4dfdf903ec8e7638dc83-.png" /> (A) (M+m) g tan θ (B) (M+m)g cot θ (C) (m/M)(M+m)g tan θ (D) (m/M)(M+m)g cot θ

Tardigrade Admin · Accepted Answer

For no relative motion between wedge and block, let the acceleration of both block and wedge be ‘a ’ towards left

From FBD of block ;

N cos θ = m g \dots (i)

and

F - N s in θ = ma \dots (ii)

From FBD of wedge;

N s in θ = M a \dots (iii)

Solving equation (i), (ii) and (iii), we get

F = \frac{m}{M} (M + m) g t an θ

$(M + m) g t an θ$

$(M + m) g co t θ$

$\frac{m}{M} (M + m) g t an θ$

$\frac{m}{M} (M + m) g co t θ$

(M+m)gtanθ

(M+m)gcotθ

Mm​(M+m)gtanθ

Mm​(M+m)gcotθ

For no relative motion between wedge and block, let the acceleration of both block and wedge be ‘a ’ towards left From FBD of block ; Ncosθ=mg…(i) and F−Nsinθ=ma…(ii) From FBD of wedge; Nsinθ=Ma…(iii) Solving equation (i), (ii) and (iii), we get F=Mm​(M+m)gtanθ

$(M + m) g t an θ$

$(M + m) g co t θ$

$\frac{m}{M} (M + m) g t an θ$

$\frac{m}{M} (M + m) g co t θ$