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Blazing goIITian

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11 Jul 2009 00:20:35 IST

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Q. Three blocks placed on top of one another on a table.there is a 7kg block.on top of it there is a

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Q. 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]

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~Áß???????

Blazing goIITian

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11 Jul 2009 13:12:31 IST

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VERY NEAT AND CLEAN ..HAHAHAHAH

ramyani chakrabarty

Blazing goIITian

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11 Jul 2009 13:17:28 IST

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Thank u.

All credit goes to swaty gupta.

ramyani chakrabarty

Blazing goIITian

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11 Jul 2009 13:21:47 IST

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And ::::

when force acts on 2kg block:

for f.b.d of 2kg block

f_max= (0.2)2.10= 4N

10 - f₁= 2a₁

a₁= 3 ms^-2

for 3kg block:

f₁- f₂= 3a₂;... (1)

for 7kg block:

f₂= 7a₃;... (2)

now, f₂ will assume the value such that the realtive motion between 3kg n 7kg block is minimum.. (friction is based on relative motion btw the surfaces)

since, f₂(max)= 15N;

a₂=a₃

so, solving these..

a₂=a₃= 0.4 ms^-2

now,

when force acts on 3kg block:

f₁= 2a₁... (1)

10 - (f₁ + f₂) = 3a₂.. (2)

f₂= 7a₃.. (3)

now, the friction adjusts itself in such a way that relative motion is least among all surfaces. so, the accelerations of blocks should be the same so that relative motion is zero.. (note that this case is not always valid.. (sometimes the friction values may not permit this.. lets try out here)

hence

a₁=a₂=a₃=a

so solving (1), (2), (3)

a=5/6 ms^-2

the case is exactly same for the third part.

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