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Mechanics

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 Joined: 19 Apr 2012 Post: 49
25 Apr 2012 11:54:15 IST
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In the arrangement shown in figure, there is friction between the blocks of masses m and 2m which are in contact. The ground is smooth. The mass of the suspended block is m. The block of mass m which is kept on mass 2m is stationary with respect to block of mass 2m. The force of friction between m and 2m is (pulleys and strings are light and frictionless) :

1)
2)
3)
4)

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Joined: 23 Jan 2007
Posts: 7958
26 Apr 2012 13:33:27 IST
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The system m+2m and other block m are connected by same string and also in the system m+2m m is stationary with respect to 2m implies acceleration of all the blocks are same.

Let the accelaration of m be 'a' downwards and that of m+2m be 'a' rightwards.

Writing force equation for m  :  mg - T = ma.....(1)

Writing force equation for m+2m  :  T = (m+2m)a........(2)

From (1) and (2)  :  a = g/4

Now consider system m+2m : block m moves with accelaration g/4 but friction with block 2m is keeping it stationary with respect to 2m and the magnitude of frictional force = ma = mg/4

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