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=2sec (in newtons)
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Solution
The FBD of the blocks are as shown
At t=2s, F=5t=5×2=10N From the FBD, we have NA=20N NB=10N Maximum value of friction force on A is fAmax=μNA=0.5×20=10N
So, at t=2s, F=fAmax Thus, the system will be at rest. Hence, fA will be equal to F i.e. 10N Therefore, T+fA=F⇒T=0