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# Help Needed here

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
2 years, 5 months ago

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Link if image not visible here · 2 years, 5 months ago

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· 2 years, 5 months ago

if it was asked to find the tension at any distance $$x$$ then we have integrate?? plz reply · 2 years, 5 months ago

Thanks for the the sourabh $$\ddot \smile$$ · 2 years, 5 months ago

No, actually don't need to . · 2 years, 5 months ago

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) · 2 years, 5 months ago

About the second problem may i speak something? · 2 years, 5 months ago

yes surely · 2 years, 5 months ago

how do i post an image? I have solved the second problem · 2 years, 5 months ago

Dont know may be satvik knows · 2 years, 5 months ago

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

! [img] (url of the image) Without spaces. · 2 years, 5 months ago

Thanks for telling the steps to upload image bro....!!!!. · 2 years, 5 months ago

You are welcome! :) · 2 years, 5 months ago

Are you guys fine? A massive earthquake has been experienced in large parts of Northern India. · 2 years, 5 months ago

I got up late in the morning so i don't have any idea of the tremors. XD · 2 years, 5 months ago

Did you sleep till 12 noon? :P · 2 years, 5 months ago

nope Till 11. :P · 2 years, 5 months ago

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. · 2 years, 5 months ago

Do we really need to solve that question using calculus when we can use the newton laws directly?? · 2 years, 5 months ago

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. · 2 years, 5 months ago

Hey @Saurabh Patil and @satvik pandey can i have ur views on this · 2 years, 5 months ago

Sorry bro, I am not very good in Maths. :( · 2 years, 5 months ago

Looks like I m late · 2 years, 5 months ago

Hey check out this problem · 2 years, 5 months ago

I got that. I solved it by using concept of pseudo force. · 2 years, 4 months ago

DO reply if u liked the problem. · 2 years, 4 months ago

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? ?? · 2 years, 4 months ago

yup i also did that using the concept of pseudo force , u should now try this one · 2 years, 4 months ago

I haven't studied Oscillations till now. :( · 2 years, 4 months ago

Thank you guyz means a lot. · 2 years, 5 months ago

Its our pleasure Right satvik pandey ???!!! · 2 years, 5 months ago

Yeah! sure! · 2 years, 5 months ago

If you know the answer then please post it. · 2 years, 5 months ago

ok answer are 1) 2.5 kg and 2) $$\frac{Mg}{2}$$ · 2 years, 5 months ago

I got 1st one. Wait I am posting. :) · 2 years, 5 months ago

@satvik pandey @Kushal Patankar @Abhineet Nayyar plz post a solution · 2 years, 5 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. :) · 2 years, 5 months ago

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. · 2 years, 5 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. :( · 2 years, 5 months ago

but u get 4kg from here but the answer is 2.5 kg · 2 years, 5 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 second · 2 years, 5 months ago