wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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)



Open in App
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

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Friction: A Quantitative Picture
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon