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 and mass is being pulled on a rough horizontal floor by a constant horizontal force . 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 . Then the tension at the mid point of the rope is
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Thanks for the the sourabh ⌣¨
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No, actually don't need to .
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if it was asked to find the tension at any distance x then we have integrate?? plz reply
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
If you know the answer then please post it.
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ok answer are 1) 2.5 kg and 2) 2Mg
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I got 1st one. Wait I am posting. :)
Thank you guyz means a lot.
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Its our pleasure Right satvik pandey ???!!!
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Yeah! sure!
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 μ coefficient of friction between A and B or B and C? Is block A fixed or movable? ??
Looks like I m late
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 Δy we get--
T(y)−T(y+Δy)−μLMΔy=LMΔy2g
limΔy→0ΔyT(y+Δy)−T(y)=−LMg
dydT(y)=−LMg
∫T(0)T(L/2)dT(y)=∫0L/2−LMg dy
T(0)=Mg
So T(2L)−T(0)=−2Mg
On putting the value of T(0)
we get T(2L)=2Mg
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+f2=mg
But there is a horizontal force acting on the block. If the block has to be at rest then N=0 So f2=0
So T=mg
Now only horizontal force acting on block of mass M is f1 and T
So T=f1
So mg=f1
As the friction is static so it can have variable magnitude. The value of m will be maximum when f1 is maximum. And the limiting value of f1 is μN0. No force is acting in vertical direction other than Normal reaction from ground and Mg. So N0=Mg
So the value of m will be maximum when mg=μMg. 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. :)
but u get 4kg from here but the answer is 2.5 kg
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.
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Hi Kushal. Thanks for pointing the mistake. I made a blunder mistake is a hurry. :( Also I substituted the wrong value of μ=1/4 that's why the the answer came out to be 2.5 kg. :(