Block A of mass m and block B of mass 2 m are placed on a fixed...

Block A of mass

\mathrm{m}

and block

\mathrm{B}

of mass

2 \mathrm{ m}

are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure. Block A of mass m and block B of mass 2 m are placed on a fixed triangular wedge by means of a massless, inextensible

Block A of mass m and block B of mass 2 m are placed on a fixed triangular wedge by means of a massless, inextensible

The wedge is inclined at

45^{{}^{\circ}}

to the horizontal on both the sides. If the coefficient of friction between the block

\mathrm{A}

and the wedge is

2 / 3

and that between the block

\mathrm{B}

and the wedge is

1 / 3

and both the blocks

\mathrm{A}

and

\mathrm{B}

are released from rest, the acceleration of A will be

Options

A $-1 \mathrm{ ms}^{-2}$
B $1.2 \mathrm{ ms}^{-2}$
C $0.2 \mathrm{ ms}^{-2}$
Dzero

Correct answer

D. zero

Step-by-step solution

Total maximum friction force,

f_{t}=f_{A}+f_{B}

(\mu_{\mathrm{A}}\mathrm{mg} \cos

45^{{}^{\circ}}+\mu_{\mathrm{B}}

\mathrm{mg} \cos

45^{{}^{\circ}}) or, (f_{t}=m g

\left(\frac{2}{3} \times \frac{1}{\sqrt{2}}+\frac{1}{3} \times

\times \frac{1}{\sqrt{2}}\right)

\frac{4}{3 \sqrt{2}} \mathrm{mg} The

net pulling force, (F=2 m g \sin 45^{{}^{\circ}}-m g \sin 45^{{}^{\circ}}) (=

\mathrm{mg} \sin

45^{{}^{\circ}}) (=

\frac{m g}{\sqrt{2}}

(\because

\mathrm{F}\lt \mathrm{f}_{\mathrm{t}}

(\therefore) The blocks will not move. So, acceleration of block A is Zero. (\therefore) option (D) is correct.

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