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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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

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

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

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

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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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Looks like I m late

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

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

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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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