1) An arrangement of pulley block system is shown above. The bigger block has mass \(10\) kg and smaller block has mass \(m\) kg. The coefficient of friction between the blocks is \(0.1\). and the coefficient of friction between the block and the ground is \(0.4\). Find the maximum value of \(m\) in kg so that the arrangement is in equilibrium.

2) A rope of length \(L\) and mass \(M\) is being pulled on a rough horizontal floor by a constant horizontal force \(F=Mg\). The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between rope and floor is \(\frac{1}{2}\). Then the tension at the mid point of the rope is

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## Comments

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img

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if it was asked to find the tension at any distance \(x\) then we have integrate?? plz reply

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Thanks for the the sourabh \(\ddot \smile\)

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No, actually don't need to .

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at all act as variable force and as we draw a FBD we see no variable force acting on the system (rope+earth)dosen'tLog in to reply

About the second problem may i speak something?

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yes surely

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how do i post an image? I have solved the second problem

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! [img] (url of the image) Without spaces.

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Are you guys fine? A massive earthquake has been experienced in large parts of Northern India.

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I got up late in the morning so i don't have any idea of the tremors. XD

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Did you sleep till 12 noon? :P

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I found a method to solve that question by using calculus.

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Applying Newton's law on \( \Delta y\) we get--

\(T(y)-T(y+\Delta y)-\mu \frac { M }{ L } \Delta y=\frac { M }{ L } \Delta y\frac { g }{ 2 } \)

\(lim\quad \Delta y\rightarrow 0\frac { T(y+\Delta y)-T(y) }{ \Delta y } =-\frac { Mg }{ L } \)

\(\frac { dT(y) }{ dy } =-\frac { Mg }{ L } \)

\(\int _{ T(0) }^{ T(L/2) }{ dT(y) } =\int _{ 0 }^{ L/2 }{ -\frac { Mg }{ L } } \) dy

\(T(0)=Mg\)

So \(T\left( \frac { L }{ 2 } \right) -T(0)=-\frac { Mg }{ 2 } \)

On putting the value of \(T(0)\)

we get \(T\left( \frac { L }{ 2 } \right) =\frac { Mg }{ 2 } \)

But I must say that Saurabh's method is better that this.

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@Saurabh Patil and @satvik pandey can i have ur views on this

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Looks like I m late

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Hey check out this problem

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I got that. I solved it by using concept of pseudo force.

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DO reply if u liked the problem.

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I recently saw your this problem. Is \( \mu\) coefficient of friction between A and B or B and C? Is block A fixed or movable? ??

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yup i also did that using the concept of pseudo force , u should now try this one

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Thank you guyz means a lot.

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Its our pleasure Right satvik pandey ???!!!

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Yeah! sure!

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If you know the answer then please post it.

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ok answer are 1) 2.5 kg and 2) \(\frac{Mg}{2}\)

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I got 1st one. Wait I am posting. :)

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@satvik pandey @Kushal Patankar @Abhineet Nayyar plz post a solution

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From the FBD of block of mass \(m\)

The block will be at rest if

\(T+f_{2}=mg\)

But there is a horizontal force acting on the block. If the block has to be at rest then \(N=0\) So \(f_{2}=0\)

So \(T=mg\)

Now only horizontal force acting on block of mass \(M\) is \(f_{1}\) and \(T\)

So \(T=f_{1}\)

So \(mg=f_{1}\)

As the friction is static so it can have variable magnitude. The value of m will be maximum when \(f_{1}\) is maximum. And the limiting value of f1 is \(\mu N_{0}\). No force is acting in vertical direction other than Normal reaction from ground and Mg. So \(N_{0}=Mg\)

So the value of m will be maximum when \(mg=\mu Mg\). Just put the values. You will get the answer. :)

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Hi satvik, You missed two forces

i.e.tension forces on the 10 kg block, one downward and other rightwards . They are acting on clamped pulley, and should be included in fbd, because clamped pulley is part of 10kg mass.Log in to reply

Hi Kushal. Thanks for pointing the mistake. I made a blunder mistake is a hurry. :( Also I substituted the wrong value of \(\mu=1/4\) that's why the the answer came out to be 2.5 kg. :(

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but u get 4kg from here but the answer is 2.5 kg

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yup i missed that thing that no horizontal force will act on the the block m , i ended up on thinking how to find \(N\), any ways thanx a lot, what about the second

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I am trying second one. :)

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