# 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
5 years, 3 months ago

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- 5 years, 3 months ago img

- 5 years, 3 months ago

Thanks for the the sourabh $\ddot \smile$

- 5 years, 3 months ago

No, actually don't need to .

- 5 years, 3 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)

- 5 years, 3 months ago

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

- 5 years, 3 months ago

About the second problem may i speak something?

- 5 years, 3 months ago

yes surely

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Dont know may be satvik knows

- 5 years, 3 months ago

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

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

- 5 years, 3 months ago

Thanks for telling the steps to upload image bro....!!!!.

- 5 years, 3 months ago

You are welcome! :)

- 5 years, 3 months ago

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

- 5 years, 3 months ago

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Thank you guyz means a lot.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Yeah! sure!

- 5 years, 3 months ago

Hey check out this problem

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

I haven't studied Oscillations till now. :(

- 5 years, 3 months ago

DO reply if u liked the problem.

- 5 years, 3 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? ??

- 5 years, 3 months ago

Looks like I m late

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Did you sleep till 12 noon? :P

- 5 years, 3 months ago

nope Till 11. :P

- 5 years, 3 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.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

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.

- 5 years, 3 months ago img

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. :)

- 5 years, 3 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

- 5 years, 3 months ago

I am trying second one. :)

- 5 years, 3 months ago

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

- 5 years, 3 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.

- 5 years, 3 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. :(

- 5 years, 3 months ago