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in Laws of motion by (15.2k points)
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Two blocks connected by a massless string slides down an inclined plane having an angle of inclination 37°. The masses of the two blocks are M1 = 4 kg and M2 = 2 kg respectively and coefficients of friction of M1 and M2 with the inclined plane are 0.75 and 0.25 respectively. Assuming the sting to be taut, find (a) the common acceleration of two masses and (b) the  tension in the string. (sin 37° = 0.6 cos 37° = 0.8) (take g = 9.8 m/s2)

2 Answers

+1 vote
by (39.0k points)
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Best answer

Maximum force of friction between M1 and inclined plane

f1 = m1M1g cos θ = (0. 75)(4)(9. 8)(0.8) = 23.52 N

M1g sin θ = (4)(9.8)(0.6) = 23.52 N = F1 (say)

Maximum force of friction between M2 and inclined plane

f2 = m2M2g cos θ = (0. 25)(2)(9.8)(0.8) = 3.92 N

M2g sin θ = (2)(9.8)(0.6) = 11.76 N = F2 (say)

Both the blocks will be moving downwards with same acceleration a. Different forces acting on two blocks are as shown in figures.

Equation of motion of M1

T + F1 – f1 = M1a     ... (i)

or T = 4a

Equation of motion M2

F2 – T – f2 = M2a

or 7.84 – T = 2a     ... (ii)

Solving eqs. (i) and (ii), we get

a = 1.3 m/s2 and T = 5.2 N

+1 vote
by (15.7k points)

a = 1.5m /s2 , T = 5.2 N

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