Derive the expression for the work done by the friction force

[Alexanddros81]

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.

Homework Equations

The Attempt at a Solution

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

 

[BvU]

$\mu$ 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.

 

[Alexanddros81]

Am going the right direction with the above calculation?

 

[haruspex]

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 N_A=P cos(φ) equation is the useful one, but you need to find P as a function of φ and integrate.

 

[Alexanddros81]

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)

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...

 

[BvU]

How would you calculate $P$ at point B ?

 

[haruspex]

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

It's the tension in a spring. What determines the value of that?

 

[Alexanddros81]

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 )

 

[PeroK]

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.

 

[Alexanddros81]

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

 

[PeroK]

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 $B$.

 

[BvU]

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 $[0,\pi/4]$

 

[Alexanddros81]

Now you need to analyse the forces at each point x between A and B

Well using Inkscape, FBD is:

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

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

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?)

 

[BvU]

$\ P_y = P \cos\phi$ is right. Both factors change during the trip. So fill in something for $P$: a function of $\phi$.

 

[Alexanddros81]

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);

 

[BvU]

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

 

[Alexanddros81]

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

 

[haruspex]

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.