Determine the magnitude of frictional force and acceleration of the block in each of the following cases :
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Solution
in Case I
N1=100N, mg=50N,
fk=μkN1=0.3×100=30N
fs=μsN1=0.4×100=40N.
As the downward force mg is greater than the force of static friction μsN1
Hence, the block will not be able to stay at rest. it will accelerate downwards. But, when it starts slipping, then kinetic friction will come into action. Now,
a=mg−fkm=50−305=4m/s2.
So, in this case, f=fk=30N and a=4m/s2(downwards).
in Case II
N2=500N, mg=500N,
fk=μkN2=0.3×500=150N
fs=μsN2=0.4×500=200N.
Here, f1 is greater than mg(drivingforce).
Hence,
Block will not move. So, in this case a=0, f=mg=50N
in case III,
N3=500cos37o=400N,
mg=120N,
fk=μkN1=0.3×400=120N
fs=μsN1=0.4×400=160N.
Here drivingforce=500sin37o−120=180N (in upward direction).
The block will move upward as the diving force is greater than the static friction.