Question

A fixed wedge with both surface inclined at ${45}^0$ to the horizontal as shown in the figure $A$ particle P of mass m is held on the smooth plane by a right string which passes over a smooth pulley $A$ and attached to a particle $Q$ of mass $3m$ which rests on a rough plane. The system is released from rest. Given that the acceleration of each particle is of magnitude $\frac{g}{5\sqrt{2}}$ then tension in the string is

• A. $mg$
• B. $\frac{6mg}{5\sqrt{2}}$
• C. $\frac{6mg}{5}$
• D. $\frac{mg}{4}$

Answer 1

The correct option is C $\frac{6mg}{5}$

$f=3\mu mg\cos \theta$

$3mg\sin \left(45\right)–3\mu mg\cos \left(45\right)–T=3ma$ -----------1.

$T–mg\sin \left(45\right)=ma$ ----------2.

Adding eqn. 1 and eqn. 2:

$2mg\sin \left(45\right)–3\mu mg\cos \left(45\right)=4ma=\frac{4mg}{5\sqrt{2}}$

$=>2–3\mu =\frac{4}{5}$

$=>\mu =0.4$

Putting value of $\mu$ in eqn 1.

$3mg\sin \left(45\right)–1.2mg\cos \left(45\right)–T=\frac{3mg}{5\sqrt{2}}$

$=>3mg–1.2mg–T=0.6mg$

$=>T=1.2mg=\frac{6mg}{5}$