Question

A fixed wedge with both surface inclined at 45° to the horizontal
as shown in the figure. A particle P of mass m is held on the
smooth plane by a light string which passes over a smooth
pulley A and attached to a particle Q of mass 3m which rests
on the rough plane. The system is released from rest. Given
that the acceleration of each particle is of magnitude then
Smooth
Rough
45° fixed 45°
the tension in the string is :
6mg
(A) mg
512
(D) mg​

Answer 1

Answer:

Explanation:

=> Suppose the coefficient of friction between the block and the rough side is equals to μ.

a = g / (5/√2)

=> According to free body diagram:

f = 3μmgcosθ  (here, θ = 45°)

3mgsin45° - 3μmgcos45° - T = 3ma ...(1)

T - mgsin45° = ma ... (2)

=> By adding eq (1) and (2), we get

2mgsin45° - 3μmgcos45° = 4ma

2 * mg* 1 /√2 - 3μ*mg * 1/√2 = 4 * m[g /5√2]

2√2 / √2 - 3√2μ/√2 = 4√2 / 5√2

2 - 3μ = 4/5

μ = 0.4

=> By Putting this value in eq (1), we get

3mgsin45° - 3μmgcos45° - T = 3ma

3mgsin45° - 3*0.4 mgcos45° - T = 3m*g/5√2

3mg - 1.2mg - T = 0.6mg

T = 1.2 mg = 6mg / 5.

Thus, the tension in the string is  6mg / 5.