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Question

Figure shows two blocks tied by a string. A variable force F=5t is applied on the block A. The coefficient of friction for the blocks A and B are 0.5 and 0.6 respectively. Find the tension in the string at time t=2 sec (in newtons)




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Solution

The FBD of the blocks are as shown


At t=2 s, F=5t=5×2=10 N
From the FBD, we have
NA=20 N
NB=10 N
Maximum value of friction force on A is
fAmax=μNA=0.5×20=10 N

So, at t=2 s, F=fAmax
Thus, the system will be at rest.
Hence, fA will be equal to F i.e. 10 N
Therefore,
T+fA=F T=0

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