Derive the expression for the work done by the friction force

In summary, the slider moves from A to B, and the coefficient of kinetic friction between the slider and the rod is μ. Deriving the expression for the work done by the friction force on the slider as it moves from A to B is necessary, but is not given in the exercise. The NA=P cos(φ) equation is the useful one, but you need to find P as a function of φ and integrate.
  • #1
Alexanddros81
177
4

Homework Statement


14.6 The coefficient of kinetic friction between the slider and the rod is μ, and the
free length of the spring is ##L_0 = b##. Derive the expression for the work done by
the friction force on the slider as it moves from A to B. Neglect the weight of the slider.

Fig P14_5_6.png


Homework Equations

The Attempt at a Solution



Pytels_Dynamics101.jpg

I[/B] am going the right direction with my solution?
I don't see to be getting the ##-0.1186μkb^2## solution.
 

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  • #2
##\mu## stands for the ratio of the friction force and some normal force. What's that normal force as a function of position ?

Pity you don't give any relevant equations. I don't think the exercise wants you to look at the velocities.
 
  • #3
Pytels_Dynamics103.jpg

Am going the right direction with the above calculation?
 

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  • #4
Alexanddros81 said:
View attachment 215969
Am going the right direction with the above calculation?
It is preferred that you type in your algebra. This makes it easier to comment on specific lines. Or you could number the equations.
Your FBD omits the force dragging the slider from A to B. You should take acceleration to be negligible. This invalidates your force balance equation for the x direction.
The NA=P cos(φ) equation is the useful one, but you need to find P as a function of φ and integrate.
 
  • #5
upload_2017-12-4_22-3-44.png


haruspex said:
Your FBD omits the force dragging the slider from A to B. You should take acceleration to be negligible. This invalidates your force balance equation for the x direction.
For the x direction:
##ΣFx=0 => F-F_k=0 => F=F_k## (1)

For the y direction:
##ΣFy=0 => N_A-Pcosφ=0 => N_A=Pcos(φ)## (2)

haruspex said:
The NA=P cos(φ) equation is the useful one, but you need to find P as a function of φ and integrate

How do I do that? I am a bit confused...
 

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  • #6
How would you calculate ##P## at point B ?
 
Last edited:
  • #7
Alexanddros81 said:
How do I do that? I am a bit confused...
It's the tension in a spring. What determines the value of that?
 
  • #8
BvU said:
How would you calculate PPP at point B ?

P at point B be should be ## -kx## where x is the elongation of the spring.

##x = L_B-L_0 = \sqrt {b^2+b^2} - b = 0.414b##

So P = -0.414kb

Am I correct? (it is the second problem with springs ever i am trying to solve )
 
  • #9
Alexanddros81 said:
P at point B be should be ## -kx## where x is the elongation of the spring.

##x = L_B-L_0 = \sqrt {b^2+b^2} - b = 0.414b##

So P = -0.414kb

Am I correct? (it is the second problem with springs ever i am trying to solve )

I think the problem is a lot more complicated than that. First, you need an expression for the length of the spring as a function of the distance from A to B.
 
  • #10
PeroK said:
I think the problem is a lot more complicated than that. First, you need an expression for the length of the spring as a function of the distance from A to B.

So to do that I split the distance between A and B to 3 sections. from 0 to 1/3 of b, from 1/3b to 2/3b, and from 2/3b to b.

At 1/3b ##sinφ = \frac {1b/3} {l} => sinφ = \frac {1b/3} {1.054b} = 0.316## (where ##l = b + L_{1/3b} - L_0##)

At 2/3b ##sinφ = 0.55##
At 3/3b ##sinφ = 0.707## that verifies that when collar is at position B the angle φ is 45 degrees.

So from above we can say that ##l = L_{AB} = \sqrt {b^2+(xb)^2}## where x is between 0 and 1
 
  • #11
Alexanddros81 said:
So from above we can say that ##l = L_{AB} = \sqrt {b^2+(xb)^2}## where x is between 0 and 1

Well, I'd say just use good old Pythagoras:

##L = \sqrt{x^2 + b^2}##, where ##0 \le x \le b##

Now you need to analyse the forces at each point ##x## between ##A## and ##B##.
 
  • #12
Alexanddros81 said:
So to do that I split the distance between A and B to 3 sections. from 0 to 1/3 of b, from 1/3b to 2/3b, and from 2/3b to b.

At 1/3b ##sinφ = \frac {1b/3} {l} => sinφ = \frac {1b/3} {1.054b} = 0.316## (where ##l = b + L_{1/3b} - L_0##)

At 2/3b ##sinφ = 0.55##
At 3/3b ##sinφ = 0.707## that verifies that when collar is at position B the angle φ is 45 degrees.

So from above we can say that ##l = L_{AB} = \sqrt {b^2+(xb)^2}## where x is between 0 and 1
Remember the last line in #4: You need ##P## as a function of ##\phi##. I think you do fine for ##\phi## at specific values, but you need a function for all ##\phi## in ##[0,\pi/4]##
 
  • #13
PeroK said:
Now you need to analyse the forces at each point x between A and B

Well using Inkscape, FBD is:

FBD P14_6.jpg


in y-axis ##N_A = P_y => N_A = Pcos(φ)##

in x-axis ##-F_k - P_x < F## (Is this correct?)

BvU said:
Remember the last line in #4: You need P as a function of ϕ. I think you do fine for ϕ at specific values, but you need a function for all ϕ in [0,π/4]

##P=Pcos(φ) ## (Is this correct?)
 

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  • #14
##\ P_y = P \cos\phi## is right. Both factors change during the trip. So fill in something for ##P##: a function of ##\phi##.
 
  • #15
BvU said:
Py=Pcosϕ is right. Both factors change during the trip. So fill in something for P: a function of ϕ
The only thing i understand is that ##P = -kl##
I can see that as the angle increases so does P. So P∝φ

should it be P related to ##1/cos(φ) ? If angle is 0deg the 1/cos(φ) is 1; Then as angle increases so does 1/cos(φ);
 
  • #16
Write it out in full so you have something that you can integrate from ##\phi=0## to ##\phi = \pi/2##...
To help you: ##k## is given. Now you need ##l(\phi)## -- you did it already, see above
 
  • #17
BvU said:
To help you: k is given. Now you need l(ϕ) -- you did it already, see above
Hi. When you say "see above" which post you are referring to?
thanks
 
  • #18
Alexanddros81 said:
Hi. When you say "see above" which post you are referring to?
thanks
Post #10, but you would do better to use PeroK's simplification in post #11. It will be more convenient to have x as the distance along the bar from A, so x ranges from 0 to b.
 

1. What is the definition of friction force?

The friction force is a resistive force that opposes the motion of an object when it comes into contact with a surface. It is caused by the microscopic irregularities on the surface of the object and the surface it is in contact with.

2. How is the magnitude of friction force determined?

The magnitude of friction force is determined by the coefficient of friction, which is a measure of the roughness of the surfaces in contact. It is also affected by the normal force, which is the force exerted by the surface on the object perpendicular to the contact surface.

3. What is the relationship between work done by friction force and the displacement of the object?

The work done by the friction force can be calculated by multiplying the magnitude of the force by the displacement of the object in the direction of the force. This means that the work done by friction force is directly proportional to the displacement of the object.

4. Can the work done by friction force be positive or negative?

The work done by friction force is always negative. This is because the friction force always acts in the opposite direction of the displacement of the object, resulting in negative work being done on the object.

5. How is the expression for work done by friction force derived?

The expression for work done by friction force is derived from the formula for work, which is force multiplied by displacement. By substituting the magnitude of friction force and the displacement of the object, the final expression for work done by friction force is obtained.

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