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Question

Two blocks of masses m1=5 kg and m2=6 kg are connected by a light string passing over a light frictionless pulley as shown in figure. The mass m1 is at rest on the inclined plane and mass m2 hangs vertically. If the angle of incline θ=30∘, what is the magnitude and direction of the force of friction on the 5 kg block. Take g=10 ms−2.

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Solution

The correct option is **B** 35 N down the plane

Weight of mass m2=6×10=60 N. The weight of m2 provides the tension. Thus

T=60 N

Opposing this force along the plane is the component F1=m1g sin θ of the force m1g. Now F1=m1g sin θ=5×10×sin 30∘=25 N. Since F1 is less than T and is, therefore, insufficient to balance T (see Fig.), the force of friction (f) down the plane is necessary to keep block m1 at rest. Thus, f must act down the plane. Since the mass m1 is at rest, the net force on m1 along the plane must be zero. Thus

T−m1g sin 30∘−f=0

or f=T−m1g sin 30∘

=60−5×10×sin 30∘

=60−25=35 N

Hence, the correct choice is (b).

Weight of mass m2=6×10=60 N. The weight of m2 provides the tension. Thus

T=60 N

Opposing this force along the plane is the component F1=m1g sin θ of the force m1g. Now F1=m1g sin θ=5×10×sin 30∘=25 N. Since F1 is less than T and is, therefore, insufficient to balance T (see Fig.), the force of friction (f) down the plane is necessary to keep block m1 at rest. Thus, f must act down the plane. Since the mass m1 is at rest, the net force on m1 along the plane must be zero. Thus

T−m1g sin 30∘−f=0

or f=T−m1g sin 30∘

=60−5×10×sin 30∘

=60−25=35 N

Hence, the correct choice is (b).

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