| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 09:19:16 IST
|
|
|
how can we solve the problems relating to the fiction and block over the block?
|
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 09:43:09 IST
|
|
|
fiction????????????????do u mean FRICTION.............???????
Pls give an example of such kind of problem so, that we can solve it for u..............
|
LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 09:49:22 IST
|
|
|
just remember that friction always opposes the relative motion between two surfaces for block over block u need to draw the free body diagram of every block post ur problem here so that we can help u out.......
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Nov 2007 10:10:12 IST
|
|
|
read HCV and trying to send you a link by elessar sir,
http://www.goiit.com/posts/list/mechanics-hcv-friction-q-22-page-98-20292.htm
|
it is not important where u stand, but in which direction u are moving |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
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/
|
|
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|