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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 13:28:08 IST
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Thanx ramayani
I will paste that post here
Three blocks placed on top of one another on a table.there is a 7kg block.on top of it there is a 3kg block on which there is a 2kg block.frictional coefficient between 2kg and 3kg is 0.2,that between 3kg and 7 kg is 0.3 and the table is smooth.find their accelerations when 10N force is applied on (a)2 kg block (b)3 kg block (c) 7 kg block.[g=10m/s2]
For the first part,
f1max = (0.2)(20)= 4 N
which isn't sufficient to keep the 2 kg block in equilibrium. So the 2 kg block shifts left and the max. value of frictional force acts.
10 - 4 = 2a1
a1 = 3m/s2
On the 3 kg block,
f2max = (0.3)(50) = 15 N
And the force acting forward is 4N. Hence it can keep the 3 kg block rest. However, the bottom block will move. Hence friction will try to prevent relative slipping.
So
a2 = a3
4 - f2 = 3a
and
f2 = 7a
a = 0.4 m/s2
Solve for the other parts similarly
![[Thumb - block.jpg]](../../../upload/2007/7/6/32ed53f71f3e37d0ff45a8c1186f6f96_11235.jpg_thumb)
For online resources in physics, visit http://phyres.blogspot.com/
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this reply: 5 points
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