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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

Note by Tanishq Varshney
1 year, 9 months ago

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Link if image not visible here Kushal Patankar · 1 year, 8 months ago

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Saurabh Patil · 1 year, 9 months ago

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@Saurabh Patil if it was asked to find the tension at any distance \(x\) then we have integrate?? plz reply Tanishq Varshney · 1 year, 9 months ago

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@Saurabh Patil Thanks for the the sourabh \(\ddot \smile\) Tanishq Varshney · 1 year, 9 months ago

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@Tanishq Varshney No, actually don't need to . Saurabh Patil · 1 year, 9 months ago

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@Saurabh Patil Actually the friction force heredosen't at all act as variable force and as we draw a FBD we see no variable force acting on the system (rope+earth) Saurabh Patil · 1 year, 9 months ago

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About the second problem may i speak something? Saurabh Patil · 1 year, 9 months ago

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@Saurabh Patil yes surely Tanishq Varshney · 1 year, 9 months ago

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@Tanishq Varshney how do i post an image? I have solved the second problem Saurabh Patil · 1 year, 9 months ago

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@Saurabh Patil Dont know may be satvik knows Tanishq Varshney · 1 year, 9 months ago

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@Tanishq Varshney First upload the picture on imgur. And then type this --

! [img] (url of the image) Without spaces. Satvik Pandey · 1 year, 9 months ago

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@Satvik Pandey Thanks for telling the steps to upload image bro....!!!!. Saurabh Patil · 1 year, 9 months ago

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@Saurabh Patil You are welcome! :) Satvik Pandey · 1 year, 9 months ago

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Are you guys fine? A massive earthquake has been experienced in large parts of Northern India. Satvik Pandey · 1 year, 8 months ago

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@Satvik Pandey I got up late in the morning so i don't have any idea of the tremors. XD Tanishq Varshney · 1 year, 8 months ago

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@Tanishq Varshney Did you sleep till 12 noon? :P Satvik Pandey · 1 year, 8 months ago

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@Satvik Pandey nope Till 11. :P Tanishq Varshney · 1 year, 8 months ago

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@Tanishq Varshney The Earthquake was experienced at 11:46 AM. I was going through Saurabh's solution at that time.

I found a method to solve that question by using calculus. Satvik Pandey · 1 year, 8 months ago

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@Satvik Pandey Do we really need to solve that question using calculus when we can use the newton laws directly?? Saurabh Patil · 1 year, 8 months ago

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@Saurabh Patil We can use Newton's Law directly. There is no problem. It's just another way of solving the same question.

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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. Satvik Pandey · 1 year, 8 months ago

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@Saurabh Patil Hey @Saurabh Patil and @satvik pandey can i have ur views on this Tanishq Varshney · 1 year, 8 months ago

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@Tanishq Varshney Sorry bro, I am not very good in Maths. :( Satvik Pandey · 1 year, 8 months ago

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Looks like I m late Kushal Patankar · 1 year, 9 months ago

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Hey check out this problem Tanishq Varshney · 1 year, 9 months ago

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@Tanishq Varshney I got that. I solved it by using concept of pseudo force. Satvik Pandey · 1 year, 8 months ago

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@Satvik Pandey DO reply if u liked the problem. Tanishq Varshney · 1 year, 8 months ago

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@Tanishq Varshney I like your problems! :)

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? ?? Satvik Pandey · 1 year, 8 months ago

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@Satvik Pandey yup i also did that using the concept of pseudo force , u should now try this one Tanishq Varshney · 1 year, 8 months ago

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@Tanishq Varshney I haven't studied Oscillations till now. :( Satvik Pandey · 1 year, 8 months ago

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Thank you guyz means a lot. Tanishq Varshney · 1 year, 9 months ago

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@Tanishq Varshney Its our pleasure Right satvik pandey ???!!! Saurabh Patil · 1 year, 9 months ago

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@Saurabh Patil Yeah! sure! Satvik Pandey · 1 year, 8 months ago

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If you know the answer then please post it. Satvik Pandey · 1 year, 9 months ago

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@Satvik Pandey ok answer are 1) 2.5 kg and 2) \(\frac{Mg}{2}\) Tanishq Varshney · 1 year, 9 months ago

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@Tanishq Varshney I got 1st one. Wait I am posting. :) Satvik Pandey · 1 year, 9 months ago

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@satvik pandey @Kushal Patankar @Abhineet Nayyar plz post a solution Tanishq Varshney · 1 year, 9 months ago

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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. :) Satvik Pandey · 1 year, 9 months ago

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@Satvik Pandey 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. Kushal Patankar · 1 year, 9 months ago

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@Kushal Patankar 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. :( Satvik Pandey · 1 year, 8 months ago

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@Satvik Pandey but u get 4kg from here but the answer is 2.5 kg Tanishq Varshney · 1 year, 9 months ago

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@Satvik Pandey 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 Tanishq Varshney · 1 year, 9 months ago

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@Tanishq Varshney I am trying second one. :) Satvik Pandey · 1 year, 9 months ago

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