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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 May 2008 11:32:41 IST
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"NORMAL FORCE ACTING ON THE BLOCK ABOUT ITS CENTRE"
IS THE QUESTION
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
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4 May 2008 11:38:46 IST
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Hold on a second guys.....the torque of normal force acting on the cubical block should be
Mg (a /2) sin alpha ....................not (1/2) Mg a sin alpha (in the understanding terms...acceleration abt the centre gets halved ..not the resultant torque ) !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
The normal force acting on the block about the centre should be M g only .
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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Doesn't matter.....
There are different cases for this one......
I'm sorry if my previous answer was wrong.. but think about this one now....
I guess the block is static...
The gravitational force acts along the centre of mass......
Frictional force produces a torque, whose magnitude can be calculated
Now the block is not rotating, that means frictional torque must be balanced, and the hero of this problem, the normal reaction, does exactly the same
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4 May 2008 11:40:04 IST
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I guess you've got me...
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4 May 2008 18:37:42 IST
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NO I HAVENT GOT HOW COME THE NORMA FORCE PRODUCE A TORQUE ABT THE CENTRE OF MASS WHEN IT IS INTERESECTING IT
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
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4 May 2008 18:56:36 IST
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ACCORDING TO THE QUESTION WHAT IS THE NORMAL FORCE
ISINT IN THE REACTION FORCE
SO EVEN IT HAS A MAGNITUDE IT WILL BE EQUAL TO MGCOS@
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
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4 May 2008 19:13:25 IST
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*ok, since the question says d block is SLIDING the net torque is 0.
*next, since its const speed so net force=0,so fricn force =mgsin@
torque due to normal force + torque due fricn =0 so magnitude of torque of normal force= 1/2mgasin@
got it??
plz rate me:)
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4 May 2008 19:15:00 IST
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You've got a wrong notion in your mind.... Normal reaction need not pass through the centre of mass...
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4 May 2008 19:15:17 IST
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also the normal force doesn't always pass thru d centre. like in, this case the normal force has 'shifted' a bit to apply a torque...
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4 May 2008 19:19:56 IST
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Just give me one reason why it must pass through the centre of mass... if it happens, you would never be able to make a turn with your bike... (if you ride one)
Normal reaction may be acting anywhere from one end to the other.....
It shifts from one point to another to balance out any net torque that is being produced, better read HC Verma - somewhere at the end of Friction or rolling I guess..... you'd find the reason why a rolling sphere ever stops....
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4 May 2008 19:23:10 IST
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@ prac.....
Normal reaction never shifts to apply a torque..... it instead shifts to balance any unbalanced torque....always tries to provide stable equilibrium to the body....
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4 May 2008 21:54:20 IST
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MY DOUBT IS THAT WHY THE TORQUE ACTS AT THE EDGE
WHY NOT BETWEEN THE CENTRE AND THE EDGE AND HOW IS THE VALUE OF N MG SIN @ NOT MG COS@
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
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5 May 2008 17:34:12 IST
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I said I was sorry for a previous reply of mine......
I was wrong when I said it acts at an edge.... it just depends on frictional torque....
The Normal reaction appears because the frictional torque tries to rotate the body... but the inclined plane doesn't allow the block to rotate... it keeps it in equilibrium....... The only force that must counter the torque produced by friction is Normal reaction and hence it must be equal to frictional torque in magnitude but must have the opposite sense....
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5 May 2008 17:36:11 IST
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Magnitude of friction is easily determined.... and the friction has only one way to act.... at the surface of the block, tangential to it....
So magnitude of frictional torque is calculated, and that's exactly equal to torque due to normal reaction
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