Given a uniform chain on an incline, find the work done by friction

In summary, the chain slides up an incline (inclination is angle 'Q') provided the hanging (vertical) part equals to 'n' times the total chain length. The work done by friction is W(by friction)=-dm gCosQ x s.
  • #1
Kanda ryu
14
1

Homework Statement


A uniform chain of mass 'm' and length 'l' rests on a rough incline (inclination is angle 'Q') with its part hanging vertically. The chain (inclined) starts moving up the incline (and the vertical part moving down) provided the hanging (vertical) part equals to 'n' times the total chain length. Find the work done by friction as it completely slides off the incline.

Homework Equations


Newton's 2nd law

The Attempt at a Solution


I assumed friction coefficient to be 'u'.
Frictional force would be N u.
I took the inclined part a system to draw the free body diagram.
N = m'gCosQ where m' is the mass of chain on the incline.
Work (fric) = N u s (displacement) x Cos 180° = - m'gCosQ x s
As mass m' will decrease as chain slides up the incline. I considered a small part of chain 'x' of mass dm. As chain is uniform density,
m/l = dm/x => dm = m x /l
For a small work dW
= -dm gCosQ dx
= - (m/l) gCosQ xdx
Let the length of vertically hanging part be 'a'
a/l = n => a=nl
So length of inclined part initially is l-nl
So I integrated xdx putting limits (l-ln) to 0 and got W= mlgCosQ u / 2 but the problem is I'm unable to find a way to write 'u' in terms of given variables.
I was hoping for an idea to find 'u' or any other better method to solve this (within the scope of basic calculus and mechanics).
 
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  • #2
Kanda ryu said:
small part of chain 'x'
That is an unusual choice and could lead to confusion. The standard approach would be to say the element is length dx and is at x from one end (which needs to be specified).
Kanda ryu said:
For a small work dW
The context is unclear. A small amount of work done on/by what and during what event? I guess you mean the work done by friction on the element as it moves its whole distance up the incline.
Kanda ryu said:
= -dm gCosQ dx
Two "d" terms looks very wrong. What is dx here?
Kanda ryu said:
Let the length of vertically hanging part be 'a'
When, at the start or at some point during the movement?
Kanda ryu said:
= - (m/l) gCosQ xdx
By some miracle, you have obtained the right expression. The element length dx has mass (m/l)dx and moves distance x up the incline.
Kanda ryu said:
putting limits (l-ln) to 0 and got W= mlgCosQ u / 2
Why does n not appear in your result?
Kanda ryu said:
I was hoping for an idea to find 'u'
You are given this for the initial state:
Kanda ryu said:
The chain (inclined) starts moving up the incline
It is not entirely clear, but I would take that to mean it started in near equilibrium.
 
  • #3
PS.. I assume the question expects you to take the tip of the incline over which the chain slides as being frictionless. More reasonably, it would have the same coefficient as anywhere else. The tension in the chain as it runs around that point would lead to a significant normal force there and consequently a lot more friction. This becomes quite complicated because the friction will mean the tension is not uniform, and will anyway depend on the acceleration. This makes for a much more advanced problem.
 
  • #4
There is a frictionless pulley at the tip of incline. The dimensions of the pulley is to be neglected.(view image)
haruspex said:
When, at the start or at some point during the movement?

At the start.

The equation I get is W(by friction) = (mlgCosQ(1-n)^2 u)/2
 

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  • #5
Kanda ryu said:
There is a frictionless pulley at the tip of incline. The dimensions of the pulley is to be neglected.(view image)At the start.

The equation I get is W(by friction) = (mlgCosQ(1-n)^2 u)/2
Ok, but now use the hint I gave you at the end of post #2 to find u.
 
  • #6
So I take the initial part of chain on incline as a system and draw it's free body diagram. The exernal forceks are normal by surface, gravitational force, friction down the incline and tension up the incline by the other part of chain. But I am finding difficulty finding friction coefficient as it seems to be cancelling each other out in the equilibrium equations. Did I mess up taking the system or making the equations?
Here are my equations;
N=MgCosQ
NCosQ + TSinQ = fSinQ + Mg
NSinQ+fCosQ=TcosQ
f+MgSinQ=T
Where N and f are normal and frictional force exerted by rough surface on inclined part of chain initially and T is tension applied by the other part of chain. As you said, I assumed the system to be in near equilibrium initially so no acceleration.
 
  • #7
Kanda ryu said:
N=MgCosQ
NCosQ + TSinQ = fSinQ + Mg
NSinQ+fCosQ=TcosQ
f+MgSinQ=T
Those four equations are all obtained by considering force balances on the sloping part of the chain. There are only two dimensions, so only two independent such equations are available. Whichever two you pick, the other two are redundant.
You need to relate the mass of that portion to the mass of the hanging portion (using n) and that to the tension. You also need to relate N to f using u.
 
  • #8
haruspex said:
. Whichever two you pick, the other two are redundant.
Ok so I picked the the equation balancing the forces pulling it up and down the incline.
Hence f + m(1-n)gSinQ = T
So tension here would be equal to weight of the part hanging so T = mng
f=m(1-n)gCosQ u
So on substituting these values in the above equation, I got;
u= [n-(1-n)SinQ]/(1-n)CosQ
And on substituting this value to the Work by friction equation;
Wf= mgl(1-n)(n-(1-n)SinQ)/2
At n=1,Wf=0 (from this equation) which is to be expected as entire chain is hanging now. I hope this equation is correct and would be glad if you confirm it.
Thanks for the support
 
  • #9
Kanda ryu said:
Ok so I picked the the equation balancing the forces pulling it up and down the incline.
Hence f + m(1-n)gSinQ = T
So tension here would be equal to weight of the part hanging so T = mng
f=m(1-n)gCosQ u
So on substituting these values in the above equation, I got;
u= [n-(1-n)SinQ]/(1-n)CosQ
And on substituting this value to the Work by friction equation;
Wf= mgl(1-n)(n-(1-n)SinQ)/2
At n=1,Wf=0 (from this equation) which is to be expected as entire chain is hanging now. I hope this equation is correct and would be glad if you confirm it.
Thanks for the support
Looks right.
 

Related to Given a uniform chain on an incline, find the work done by friction

1. What is the definition of work done by friction?

Work done by friction is the force of resistance that is generated when two surfaces slide against each other. It is measured by the product of the frictional force and the distance over which the surfaces move.

2. How is the work done by friction calculated?

The work done by friction can be calculated by multiplying the magnitude of the frictional force by the distance over which the surfaces move. This can be represented by the equation W = Ff * d, where W is the work done, Ff is the frictional force, and d is the distance.

3. Why is the work done by friction important in understanding the motion of objects on an incline?

The work done by friction is important in understanding the motion of objects on an incline because it is one of the main forces that affects the motion of an object. It is responsible for slowing down or stopping the object's motion and plays a crucial role in determining the object's acceleration and velocity.

4. How does the coefficient of friction affect the work done by friction?

The coefficient of friction, which is a measure of the amount of resistance between two surfaces, directly affects the work done by friction. A higher coefficient of friction means a greater resistance and therefore a larger work done by friction.

5. Can the work done by friction ever be negative?

Yes, the work done by friction can be negative if the object is moving in the opposite direction of the frictional force. This means that the friction is actually assisting the object's motion rather than resisting it.

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