on the second body friction by the upper body froward n by lower body backward
on the third body friction by ground backward n by the upper body forward
then write newtons 2nd law u ll getthe solution
if u r still nt satisfied nudge book me
Let the friction force between the 5 and 10 kg blocks be f1, between 10 and 15 kg block be f2 and between 15 kg and the ground be f3.
f1max = 0.2 x 5 x 10 = 10 N
f2max = 0.2 x 15 x 10 = 30 N
f3max = 0.5 x 30 x 10 = 150 N
The upper block can't stay in equilibrium as the applied force is greater than the maximum value of the frictional force. The net rightward force on the second block is 100 + 10 =110 N, which is greater than the maximum value of the frictional force between the 2nd and third block. So the second block too will accelerate. However, the net force on the bottommost block is 30 N rightwards, which is lesser than the maximum frictional force between the 3rd block and the ground. Hence the 3rd block remains at rest.
a1=(50-10)/5 = 8
a2=(110-30)/10 = 8
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