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Cool goIITian

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31 Aug 2009 19:32:41 IST

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please solve: minimum force to pull up a mass on incline (with a spring)

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Please solve this problem for me ................. Thanks

1] A block of mass 'm' is attached with a massless spring of force constant 'k'. The block is placed over a rough inclined surface for which the coefficient of friction is 3/4. The minimum value of 'M' required to move the block up the place is ........

[A] 3m/5 [B] 6m/5 [C] 2m [D] m/5.

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answer given is: 3m/5

*[Note: sin37 = 3/5, cos37 = 4/5, tan37 = 3/4].

* this note is NOT given in the problem.

please solve this for me.

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rumki chakraborty

Cool goIITian

Joined: 10 Oct 2007

Posts: 41

31 Aug 2009 22:34:49 IST

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From the figure itself, it's clear that kx=T.

Forming the different equations, we have

$Mg\;-\;T\;=\;Ma$

Since the block just has to move up, i.e., it just has to start its motion, so we'll consider the acceleration to be zero.
So, now we have

$Mg\;=\;T$

Similarly, we'll get the other equation as

$T\;-\;mg\;sin\;37\;-\;\mu mg\;cos\;37\;=\;0$

T=Mg (Just shown). Put the different values and get the answer as:-

$M\;=\;\frac{6}{5}m$

but the answer given is : 3m/5.

this problem is given at four locations and all with the same answer: 3m/5

Manu Gupta

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Joined: 23 Aug 2009

Posts: 27

1 Sep 2009 19:17:10 IST

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Hello Friend,

There are a few things to understand the concept involved here.

1) The block will move when the force 'kx' exceeds the gravitational component and frictional force.

that is,

in limiting condition,

kx= mg sin37 + (3/4)mg cos37..........where 3/4 is frictional coefficient.

2) now we have to keep in mind that the force "kx" is DEPENDENT on the motion.

so, lets imagine this situation physically.

Initially, we assume, that small block m is kept and we start attaching weights of M to the string on the right side.

When M is very small, the string will move for a small amount downwards, let us call it dL. Thus, the spring force would be k*dL

BUT, how do we know the value of "dL".

There comes the concept of energy.

potential energy lost by mass M in coming down a vertical distance "dL" is Mg*dL, and this is also equal to the potential energy stored in spring with the extension of same length dL.

Thus, we can write,

Mg*dL = 1/2 * k* (dL)^2..........................spring potential energy= gravitational potential energy.

so, we get,

Mg= (k*dL)/2...........THIS IS IMPORTANT...for what I have seen in your solution, this was the point of error.

Now, Tension T which is equal to k*x, is actually double of M*g.....

After this correction, you have already solved the problem.

So, following the similar steps, we get the correct answer.

GOOD LUCK!

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