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Mechanics

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21 Sep 2007 21:25:22 IST

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Extraordinarily Conceptual !!!

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Spring fitted doors close by themselves when released.You want to keep the door open for a long time say 1hr.

if you put a half kg stone in front of the open door ,it does not help. The stone slides with the door and the door

gets closed.However if u sandwich 20g pice of wood in the small gap between the door and the floor,the door stays open.

Explain why a much lighter piece of wood is able to keep the door open while the heavy stone fails..

Please explain with help of suitable forces.

Please give sensible answers, since i have had many stupid answers for my previous quests.

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Gaurav |spideyunlimited| Ragtah

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21 Sep 2007 21:48:06 IST

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as the door nears its normal closed position , the spring is returning to its normal position, thus the oppoistion to the placed stone decreases if the door is less open. the spring is less stretched when the small stone is placed hence it provides less force, while with the large stone the spring is quite stretched at tht position and hence as it tries to return to its unstretched position with greater force, it slides away the large stone.

*PLZ RATE* thnks

James Bond

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21 Sep 2007 22:06:56 IST

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Here's the pic...

Shikhar Shukla

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22 Sep 2007 09:12:15 IST

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when the 1kg block is placed in front of the door,its normal rxn is10 N(wich is low).And when 20 g block is fixed below the door,it experiences a pressure on the block leading to an excessive increase in its normal rxn.as frictional force increses with norm rxn,friction will be strong enough (in 2nd case)to be able to stop the door against spring force.

moreover,if the 1kg block is placed towards the farther end(away frm the hinge),it may create a torque equal to that of the spring so as to prevent closing of the door.

edison

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22 Sep 2007 10:49:31 IST

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When a stone of say 1/2 kg is kept against the door then the forces acting on the door are

1) Restoring force of the spring that tends door to close = K x
(here K is the spring constant, 'x' is the extension in spring)
2) Frictional force due to the sliding of stone that opposes force due to spring, F = (coeff. of friction) * Weight of stone = u * mg = u*g

Thus here Kx > u*g , and thus the door is still not prevented from getting closed, as the frictional force of the stone is not sufficient to balance the force due to spring.

Now in the SECOND case when say 20g of piece is inserted between the gap the, reaction force (due to the wight of door and other rigid structures on the gap is so large that the normal reaction force ' N' by Piece on the floor excceds tremendously.

Thus frictional force is now f = u* N
and u* N = K x , which prevents the door from closing. As you increase the force on the door to close it the Normal reaction force 'N' increases in sucha manner so that the FORCES ARE IN EQUILIBRIUM.

THE PROBLEM IS SIMILAR TO THE FOLLOWING EXAMPLE
WHEN A BLOCK OF MASS 'M' IS KEPT ON SOME FLOOR WITH COEFFICINT OF FRICTION BEING 'u'. IF A FORCE IS APPLIED HORIZONTALLY ON THE BLOCK IT SLIDES ON THE FLOOR. NOW IF YOU KEEP INCREASING THE ANGLE OF THE FORCE WITH THE HORIZONTAL THEN AT SOME ANGLE @ OR GREATER THAN @ THE FORCE WILL NOT BE ABLE TO MOVE THE BLOCK EVEN IF THE FORCE IS INCREASED BEYOND THAT.

THIS IS AGAIN DUE TO THE FACT THAT AT ANGLE @ THERE IS A VERTICAL COMPONENT OF FORCE F IN THE DOWNWARD DIRECTION AND THUS NORMAL REACTION FORCE DUE TO THE BLOCK BECOMES N = ( M + F Sin@)

SO FRICTIONAL FORCE IS f = u * N = u * ( M + F Sin@)

MOREOVER, F Cos@ < u * ( M + F Sin@)

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