Find the accelerations a1, a2 , a3 of the three blocks shown in figure (6-E8). If a horizontal force of 10N is applied on (a) 2 kg block (b) 3 kg block , (c) 7 kg block. Take g= 10 m/s².Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "
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Solution :
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μ1=0.2
μ2=0.3
μ3=0.4
When force applied is 10N is applied on 2kg block, it experience maximum frictional force.
μR1=μx2g=0.2x20=4N from the 3kg block
so the 2kg block experience a net force =10-4=6N
So, a1=6/2=3m/s2
But for the 3kg block, the frictional force from 2kg block [4N] becomes the driving force and the maximum frictional force between 3kg and 7kg blocks.
μ2R2=0.3x5kg=15N
but for the 3kg block cannot move relative to the 7kg block , the 3kg block and the 7kg block both have same acceleration. a2=a3
which will be due to the 4N force because there is no frictional force from the floor.
a2=a3=4/10=0.4m/s2
b] when the 10 N force is applied to the 3kg , it can experience maximum frictional force of 15 +4=19N from the 2kg block and 7kg block.
so it cannot move with respect to them.
as the floor is fricitionless, all the three bodies will move together.
∴a1=a2=a3=10/12=5/6 m/s2
c] similarly we can say that when 10N force is applied to 7kg block , all three blocks will move together.
again a1=a2=a3=5/6 m/s2
*******************************************
μ1=0.2
μ2=0.3
μ3=0.4
When force applied is 10N is applied on 2kg block, it experience maximum frictional force.
μR1=μx2g=0.2x20=4N from the 3kg block
so the 2kg block experience a net force =10-4=6N
So, a1=6/2=3m/s2
But for the 3kg block, the frictional force from 2kg block [4N] becomes the driving force and the maximum frictional force between 3kg and 7kg blocks.
μ2R2=0.3x5kg=15N
but for the 3kg block cannot move relative to the 7kg block , the 3kg block and the 7kg block both have same acceleration. a2=a3
which will be due to the 4N force because there is no frictional force from the floor.
a2=a3=4/10=0.4m/s2
b] when the 10 N force is applied to the 3kg , it can experience maximum frictional force of 15 +4=19N from the 2kg block and 7kg block.
so it cannot move with respect to them.
as the floor is fricitionless, all the three bodies will move together.
∴a1=a2=a3=10/12=5/6 m/s2
c] similarly we can say that when 10N force is applied to 7kg block , all three blocks will move together.
again a1=a2=a3=5/6 m/s2
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Answer:
From the free body diagram, we get
T+0.5a-0.5g=0-----------------(1)
μR +1a+T1 -T=0-----------------(2)
μR+1a-T1=0
μR +a= T1------------(3)
From equations i , ii and iii
μR+a=T-T1
T-T1=T1
T=2T1
Equation II becomes μR +a+T1-2T1=0
μR+a-T1=0
T1=μR+a
T1=0.2g +a
Equation I will become
2T1+0.5a-0.5g =0
T1=0.5g -0.5a/2
=0.25g-0.25a
From equation Iv and V
0.2g +a=0.25g -0.25 a
a=0.05/1.25 x10
a=0.4 x 10
a=0.4 m/s2
Therefore acceleration of 1kg block is 0.4m/s2
b) Tension T1=0.2g +a+0.4=2.4 N
c) t=0.5g-0.5a
t=0.5x10-0.5x0l4
t=4.8N
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