Question

Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in Figure. The wedge is inclined at 45o to the horizontal on both the sides. The coefficient of friction between block A and wedge is 2/3 and that between block B and the wedge is 1/3. If the blocks A and B released from rest, find 

(A) the acceleration A, 

(B) Tension in the string and 

(C) The magnitude and direction of the force of friction acting on A

Answer 1

(a) Acceleration of block A

Maximum friction force that can be obtained at A is

Therefore, maximum value of friction that can be obtained on the system is

Therefore, the system will not move or the acceleration of block A will be zero.

(b) and (c) Tension in the string and friction at A

Net pulling force on the system (block A and B)

F = F₁ – F₂ = mg/√2

Therefore, total friction force on the blocks should also be equal to mg/√2

or f_A + f_B = F = mg/√2

Now since the blocks will start moving from block B first (if they move), therefore, f_B will reach its limiting value first and if still some force is needed, it will be provided by f_A

Here, (f_max) B < F

Therefore, f_B will be in its limiting value and rest will be provided by f_A.

The FBD of the whole system will be as shown in the figure

Therefore, friction on A is

f_A = mg/3√2 (down the plane)

Now for tension T in the string, we may consider either equilibrium of A or B

Equilibrium of A gives

Answer 2

Correct answer is

(a) 0, 

(b) T = 2 \(\sqrt{2}\) mg / 3 

(c) f = mg / 3 \(\sqrt{2}\) (down the plane)