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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Link if image not visible here – Kushal Patankar · 1 year, 6 months ago

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– Tanishq Varshney · 1 year, 6 months ago

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

– Tanishq Varshney · 1 year, 6 months ago

Thanks for the the sourabh \(\ddot \smile\)Log in to reply

– Saurabh Patil · 1 year, 6 months ago

No, actually don't need to .Log in to reply

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, 6 months agodosen'tLog in to reply

About the second problem may i speak something? – Saurabh Patil · 1 year, 6 months ago

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– Tanishq Varshney · 1 year, 6 months ago

yes surelyLog in to reply

– Saurabh Patil · 1 year, 6 months ago

how do i post an image? I have solved the second problemLog in to reply

– Tanishq Varshney · 1 year, 6 months ago

Dont know may be satvik knowsLog in to reply

! [img] (url of the image) Without spaces. – Satvik Pandey · 1 year, 6 months ago

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– Saurabh Patil · 1 year, 6 months ago

Thanks for telling the steps to upload image bro....!!!!.Log in to reply

– Satvik Pandey · 1 year, 6 months ago

You are welcome! :)Log in to reply

Are you guys fine? A massive earthquake has been experienced in large parts of Northern India. – Satvik Pandey · 1 year, 6 months ago

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– Tanishq Varshney · 1 year, 6 months ago

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

– Satvik Pandey · 1 year, 6 months ago

Did you sleep till 12 noon? :PLog in to reply

– Tanishq Varshney · 1 year, 6 months ago

nope Till 11. :PLog in to reply

I found a method to solve that question by using calculus. – Satvik Pandey · 1 year, 6 months ago

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– Saurabh Patil · 1 year, 6 months ago

Do we really need to solve that question using calculus when we can use the newton laws directly??Log in to reply

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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, 6 months ago

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@Saurabh Patil and @satvik pandey can i have ur views on this – Tanishq Varshney · 1 year, 6 months ago

HeyLog in to reply

– Satvik Pandey · 1 year, 6 months ago

Sorry bro, I am not very good in Maths. :(Log in to reply

Looks like I m late – Kushal Patankar · 1 year, 6 months ago

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Hey check out this problem – Tanishq Varshney · 1 year, 6 months ago

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– Satvik Pandey · 1 year, 6 months ago

I got that. I solved it by using concept of pseudo force.Log in to reply

– Tanishq Varshney · 1 year, 6 months ago

DO reply if u liked the problem.Log in to reply

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, 6 months ago

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one – Tanishq Varshney · 1 year, 6 months ago

yup i also did that using the concept of pseudo force , u should now try thisLog in to reply

– Satvik Pandey · 1 year, 6 months ago

I haven't studied Oscillations till now. :(Log in to reply

Thank you guyz means a lot. – Tanishq Varshney · 1 year, 6 months ago

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– Saurabh Patil · 1 year, 6 months ago

Its our pleasure Right satvik pandey ???!!!Log in to reply

– Satvik Pandey · 1 year, 6 months ago

Yeah! sure!Log in to reply

If you know the answer then please post it. – Satvik Pandey · 1 year, 6 months ago

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– Tanishq Varshney · 1 year, 6 months ago

ok answer are 1) 2.5 kg and 2) \(\frac{Mg}{2}\)Log in to reply

– Satvik Pandey · 1 year, 6 months ago

I got 1st one. Wait I am posting. :)Log in to reply

@satvik pandey @Kushal Patankar @Abhineet Nayyar plz post a solution – Tanishq Varshney · 1 year, 6 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, 6 months ago

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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, 6 months agoLog in to reply

– Satvik Pandey · 1 year, 6 months ago

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. :(Log in to reply

– Tanishq Varshney · 1 year, 6 months ago

but u get 4kg from here but the answer is 2.5 kgLog in to reply

– Tanishq Varshney · 1 year, 6 months ago

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 secondLog in to reply

– Satvik Pandey · 1 year, 6 months ago

I am trying second one. :)Log in to reply