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Question

Masses M1,M2 and M3 are connected by strings of negligible mass which pass over massless and frictionless pulleys P1 and P2 as shown in the figure. The P1 and P2 is parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37o with the horizontal. If the mass M1 moves downwards with a uniform velocity find
(i) the mass of M1
(ii) the tension in the horizontal portion of the string. (g=9.8 m/s2 and sin37o3/5)
1010339_959a2b8f2fbd497ba2ec77519476e146.png

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Solution

Since, the masses are moving with constant velocity, there is no acceleration and hence no external force is acting on them.

Force balance:

For mass m1:

T1=m1g (i)

Formassm3

T2=μm3g

T2=0.25×4×10

T2=10N

Formassm2:

T1=m2μgsinθ+m2gcosθ+T2

T1=4×0.25×10×35+4×10×45+T2

T1=6+32+10

T1=48N

Puttingin eqn(i)

48=m1×10

m1=4.8kg

Therefore,

The mass m1=4.8 kg and the tension in the horizontal spring T2=10N

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