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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.
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
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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
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Verified Answer
The correct answer is:
zero
Total maximum friction force,\(f_{t}=f_{A}+f_{B}\)
\(\mu_{\mathrm{A}} \mathrm{mg} \cos 45^{\circ}+\mu_{\mathrm{B}} 2 \mathrm{mg} \cos 45^{\circ}\)
or, \(f_{t}=m g\left(\frac{2}{3} \times \frac{1}{\sqrt{2}}+\frac{1}{3} \times 2 \times \frac{1}{\sqrt{2}}\right)\)
\(=\frac{4}{3 \sqrt{2}} \mathrm{mg}\)
The net pulling force, \(F=2 m g \sin 45^{\circ}-m g \sin 45^{\circ}\) \(=\mathrm{mg} \sin 45^{\circ}\)
\(=\frac{m g}{\sqrt{2}}\)
\(\because \mathrm{F}<\mathrm{f}_{\mathrm{t}}\)
\(\therefore\) The blocks will not move.
So, acceleration of block A is Zero.
\(\therefore\) option (D) is correct.
\(\mu_{\mathrm{A}} \mathrm{mg} \cos 45^{\circ}+\mu_{\mathrm{B}} 2 \mathrm{mg} \cos 45^{\circ}\)
or, \(f_{t}=m g\left(\frac{2}{3} \times \frac{1}{\sqrt{2}}+\frac{1}{3} \times 2 \times \frac{1}{\sqrt{2}}\right)\)
\(=\frac{4}{3 \sqrt{2}} \mathrm{mg}\)
The net pulling force, \(F=2 m g \sin 45^{\circ}-m g \sin 45^{\circ}\) \(=\mathrm{mg} \sin 45^{\circ}\)
\(=\frac{m g}{\sqrt{2}}\)
\(\because \mathrm{F}<\mathrm{f}_{\mathrm{t}}\)
\(\therefore\) The blocks will not move.
So, acceleration of block A is Zero.
\(\therefore\) option (D) is correct.
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