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Q.

Two blocks of masses 1 kg and 0.75 kg are attached through a massless string fixed on a  triangular wedge 

\large {\mu _1} = \frac{1}{{\sqrt 3 }}{\text{ }}and{\text{ }}{\mu _2} = \frac{2}{{\sqrt 3 }}

  Which of the following is correct?

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a

System is in equilibrium

b

Block of mass 1 kg moves upward with acceleration 3.14 m/s2

c

Block of mass 1 kg moves downward with acceleration 3.13 m/s2

d

Both block moves with same acceleration 3 m/s2

answer is A.

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Detailed Solution

 

{f_{{L_1}}} = {\mu _1}mg\cos \theta
= \frac{1}{{\sqrt 3 }} \times 1 \times 10 \times \frac{1}{2} = \frac{5}{{\sqrt 3 }} = \frac{5}{{1.7}} \equiv 3\,
{f_{down}} = mg\sin \theta
= 1 \times 10 \times \frac{{\sqrt 3 }}{2} = 5\sqrt 3 = 5 \times 1.73 = 8.6
So\;body\;{f_{{L_2}}} = \mu mg\cos \theta
= \frac{{\cancel{2}}}{{\sqrt {\cancel{3}} }} \times \frac{3}{4} \times 10 \times \frac{{\sqrt {\cancel{3}} }}{{\cancel{2}}} = 7.5\;N
{(mg\sin \theta )_2} = \frac{3}{4} \times 10 \times \frac{1}{2} = \frac{{30}}{8} = 3.75\;N

clearly system tries to move left side downward.

The net driving force acting on the system due to gravity = (8.6 - 3.75) N = 4.85 N

Net frictional force which opposes the tendency of sliding = (3 + 7.5) N = 10.5 N

Hence, the system will be in equilibrium.

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Two blocks of masses 1 kg and 0.75 kg are attached through a massless string fixed on a  triangular wedge   Which of the following is correct?