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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

In this figure, mass M is fixed on the wall.How much force on mass m=5 kg should be applied so that it won't slip.

(given: acceleration of van=5m/s^2 ;coefficient of friction between m and M=0.4)

Pls help.

*Last edited by niharika_kumar (2014-08-22 04:48:09)*

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The reaction force between m and M is 5kg*5m/s^2

To balance the friction, you must give a force of 5kg*5m/s^2*0.4

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

I thought the the same but it is an MCQ question with options as follows:-

a)170 N

b)180 N

c)150 N

d)250 N

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**Olinguito****Member**- Registered: 2014-08-12
- Posts: 648

Weight of m (taking ) is .So the frictional force required to keep m from slipping is .This can be supplied by a horizontal force of .Now the van is accelerating to the right, meaning m would be accelerating to the left relative to the van with a force of .Therefore the horizontal force required to pin m to M is .

NB: The 150 N horizontal force consists of 25 N required to keep m stationary relative to the van and 125 N needed to generate the frictional force to balance its weight.

*Last edited by Olinguito (2014-08-22 23:39:15)*

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

wow, you explained it so easily and well.(It reminded me of bob.)

thank you

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