Question

Example 5.9 What is the acceleration of
the block and trolley system shown in a
Fig. 5.12(a), if the coefficient of kinetic friction
between the trolley and the surface is 0.042
What is the tension in the string? (Take g =
10 m s?). Neglect the mass of the string.​

Answer 1

Kinetic coefficient of friction μk=0.04 given. From the diagram the equations of motion of the system network is given by,30−T= M1a or 30−T=3×a −−−−−(1) andT−fk=M2a−−−−−−(2).In the above equations M1 and M2 are the masses of block and trolley.fk=μkR and R=M2×g=20×g=20×10=200N.Therefore fk=0.04×200=8N. So equation (2) becomes,T−8=20×a−−−−(3).Adding equations (1) and (3),30−8=23×a. or a=2223. or a=0.956 ms−2.And T= 20×a+8. Or T=20×2223+8=440+18423=62423. Or T=27.1 NTherefore acceleration a=0.956 ms−2 and tension T= 27.1 N.

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