Find the friction force between block A and the surface for system at rest given below: mA=9kg,mB=4kg,μs=μk=0.8
A
72N
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B
30N
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C
40N
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D
48N
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Solution
The correct option is C40N From the FBD of Block A, to remain at rest: T=f...(i) N=mAg=90N...(ii)
From FBD of block B, to remain at rest: T=mBg=40N...(iii) ∴T=f=40N
Now fsmax=μsN=0.8×90=72N ∵f≤fmax, hence the frictional force will be equal to applied for, f=40N will act on block A.