Classical Mechanics-Inclined plane tricky problem

[squareroot]

Homework Statement

An inclined plane is given , inclined by the angle β.A small object with a given mass m is placed at the top.What distance and in what time will it go if the friction coefficient goes by the law mu=bx, where b is a constant and x is the distance that the object has traveled.

Homework Equations

All the equations from classical mechanics.

The Attempt at a Solution

I've scorched my brain for this one , but I can't find the solution , it's been a week now since I got the problem and i think it's time i ask for some help.

Thanks!

 

[golanor]

Hello!
What have you tried to do so far?
Usually, when friction is dependent on velocity/distance, it means you will need to solve a differential equation.
Where exactly did you get stuck trying to solve this?

 

[squareroot]

I'm simply stuck in the math.I have no ideea how to start solving it , i haven't studied differential equations yet.(i'm in the 11'th grade)My teacher gave me this problem so i think there is a simpler way to solve it involving only the knowledge that i have so far.In math , this is my first year of calculus studying limits and strings of numbers.
I need an idea to start with.Boy am I stuck with this one!

 

[golanor]

The best way to start is going through the basics.
Draw the issue at hand, chart all the forces, and think about this:
You know at what speed it started moving. You know at what speed it stopped moving.
What can you do with that knowledge?
Be sure to choose and easy coordinate system so the problem will be simpler to understand with the formulas.

 

[squareroot]

I think I solved the distance problem , so writing Newton's law for the moment when the object stops moving.
F=ma , where a=0 so F=0.(1)
But also F=mgsinβ-MU*N=mgsinβ-b*dmgcosβ (2)

From (1)(2) we have that mgsinβ=bdmgcosβ , where the x from MU=bx becomes the maximum distance covered by the object ,the one that i need and N is equal to the component of G that is mgcosβ.
From here we get that d=sinβ/bcosβ.

Is it correct?

for the second point , finding the time , now this i think is a bit harder, i could use it from the motion law t=√2d/a , but finding a could be a problem.

 

[squareroot]

a little update for the time , we know that F=ma so a=F/m but F=mg(sinβ-mucosβ) and t=√2x/a

so t=√2x/(g(sinβ-bxcosβ)) , is it correct like this?basically I found a formula that let's you calculate the time to any given distance x.But my question is , is this what the problem asks?

 

[golanor]

Regarding the distance, that's correct. You should try to check it yourself and see if it makes sense. What are the units of b? What should d's units be?
You can also try and test your answer by going through a different way. If you have the distance traveled and the angle, you can use the work-energy theorem to see if it works out with what you found.
The time part, however, isn't right, since when it stops a=0, and you will have t=infinity, which probably isn't the answer you are looking for. Honestly, I can only see a way to find t which you cannot use, so i'll try and think about a simpler way to do it.

 

[squareroot]

Does anyone have any idea how to solve the time problem?Ty.

 

[haruspex]

I think I solved the distance problem , so writing Newton's law for the moment when the object stops moving.
F=ma , where a=0 so F=0.(1)

No, that's when it stops accelerating. It will have some residual KE to burn off still.
Can you write an expression for the energy expended overcoming friction in traveling a distance x down the plane?

 

[squareroot]

tu,but i talked to my teacher and he told me to ignore that part,the ine with the KE.my only trouble now

 

[squareroot]

is with the time now.

 

[haruspex]

tu,but i talked to my teacher and he told me to ignore that part,the ine with the KE.my only trouble now is with the time

That's rather surprising. I can figure out the stopping distance from conservation of energy, but I see no way to get the stopping time without using differential equations. (It's basically simple harmonic motion.)

 

[golanor]

That's rather surprising. I can figure out the stopping distance from conservation of energy, but I see no way to get the stopping time without using differential equations. (It's basically simple harmonic motion.)

Same here, I tried to ask others, no one seems to find a way without using differential equations.

 

[haruspex]

squareroot, despite what you teacher side, I would recommend trying to get the right answer for the distance. It's not hard.

 

[squareroot]

i got a big hint from my techear, he said that there is a very simple way of solving this and avoid differential equations:), by finding a simillarity between this problem and mehanical oscillations...he said that if you observ that simillarity the prblem solves very easy...

 

[haruspex]

i got a big hint from my techear, he said that there is a very simple way of solving this and avoid differential equations:), by finding a simillarity between this problem and mehanical oscillations...he said that if you observ that simillarity the prblem solves very easy...

I did mention that it was basically SHM, but here there is a kind of forcing term. To take the standard SHM formulae and assume they apply would be distinctly unrigorous, but if your teacher has sanctioned it...
So, what can you write down for the period, treating this as SHM?

 

[squareroot]

the period is T=2pisqrt(m/k)

 

[haruspex]

the period is T=2π√(m/k)

Of course, but what is k in this context? Without the differential equation, it's not obvious. When you have determined the period, T, what do you think the stopping time will be?

 

[squareroot]

well i won t be studying differential equation 3 years from now so this is a bit hard for me.
Does k=MU?
As far as what the stopping time is , i have no idea.

 

[squareroot]

A, as for the time. does 2t=T? because T is the time of a complete oscillation and the body will have to go all the way down and back up so that we can say that he completed a full oscillation.

 

[haruspex]

well i won t be studying differential equation 3 years from now so this is a bit hard for me.
Does k=MU?

No. With a spring, the restoring force is kx. What is the force up the slope here?

As far as what the stopping time is , i have no idea.

The mass starts from rest. How long does it take a pendulum to go from being momentarily at rest to being so again?

 

[squareroot]

well the force is the tangential gravity minius the friction force,mgsinalpha-muN=mg(sinalpha-bxcosalpha)

 

[squareroot]

but what do i do from here?

 

[haruspex]

well the force is the tangential gravity minius the friction force,mgsinalpha-muN=mg(sinalpha-bxcosalpha)

Right. Here, in the absence of a proper ODE treatment, you have to take it on trust that the relevant part will be the coefficient of x in that force. So ignore the constant component. Comparing this with the restoring force from a spring, what equates to k?

 

[squareroot]

that s exactly what i can t figure out!

 

[haruspex]

Treating the friction up the plane as the 'restoring force' (because the downplane component of gravity is constant), it's mgbxcos(β). The restoring force in a spring is kx. If we're to regard the one as equivalent to the other, what is k equal to?

 

[squareroot]

ok, so i have the period now,can i write that time t=1/4T ?because the body moves only quarter in the time t:)

 

[haruspex]

ok, so i have the period now,can i write that time t=1/4T ?because the body moves only quarter in the time t:)

Yet again, without the reassurance of the differential equations it's unclear what fraction of a period the time to stopping represents. It turns out that the way to think of it is that its speed is 0 initially and you want the time until its speed is zero again.

 

[squareroot]

All i know is that there is a way to solve this without using differential equations , and that's for sure.

Another tip from my teacher, the restoring force is not just bxmgcoslapha , he said that i need to find a way to write this entire expression mgsinalpha-bxmgcosalpha unter the form kx , the entire expression not just the friction force.

 

[haruspex]

the restoring force is not just bxmgcoslapha , he said that i need to find a way to write this entire expression mgsinalpha-bxmgcosalpha unter the form kx , the entire expression not just the friction force.

True. It just takes a change of variable. I.e. you have to measure x from a different origin. Can you see how to do this?

 

[squareroot]

No.I'm really stuck with this one, i think I'm having the biggest mind block of my life so far.

 

[haruspex]

No.I'm really stuck with this one, i think I'm having the biggest mind block of my life so far.

What value of x makes mgsin(α)-bxmgcos(α) = 0?
We want to relocate the origin so as to be a measure of distance from that point.

 

[squareroot]

Well that would mean that mgsin(α)=bxmgcos(α) so do I write x=mgsin(α)/bmgcos(α) ?

"We want to relocate the origin so as to be a measure of distance from that point." I don t really get the meaning of this part.

 

[squareroot]

But if i take that expression and simplify it i get that x=sin(α)/bcos(α), and I've got to this expression earlier but my teacher said it s not good.

 

[haruspex]

Well that would mean that mgsin(α)=bxmgcos(α) so do I write x=mgsin(α)/bmgcos(α) ?

No, you want to shift the origin of x by that much. I.e. instead of using x as the measure of displacement you use, say, u = x - tan(α)/b. The variable u should then satisfy the standard form for SHM.

 

[squareroot]

and where woud I use this u? this is a optional problem given by my teacher for those who want a deeper grasp of shm, i need it solved and then i need too study the solving method so that i get a deep understanding, not just fractals and pieces of information.If someone could solve this and explain it to me i think i woud have a better grasp of it.As i said this is a optional problem it s not worth any credits so it s not cheating, just satisfying my curiosity.

Thank you!

 

[haruspex]

and where woud I use this u?

All this has done is measure x from a different point on the plane. Or to put it another way, we found the equilibrium position, which was not x=0. It is u=0.

 

[squareroot]

well yea, but what do i do with this u? and why is the equillibrium position at u=0?

 

[haruspex]

well yea, but what do i do with this u?

You don't have to anything with it. Reformulating in terms of this u allows you to recognise the equation as being an exact fit for SHM. That allows you to figure out the frequency and amplitude in the usual way. Your problem was that extra constant term in the acceleration; rewriting it with u got rid of that.

and why is the equillibrium position at u=0?

Because that is where the acceleration will be 0.

 

[squareroot]

got it! but where do i use this u in equations?that s what i can t figure out.

 

[squareroot]

and why if i write it u like that, why does it the resemble a smh?

 

[squareroot]

what i want to say is how do i find k? with this u?

 

[haruspex]

Measuring x from the top you had:
F=mgsin(β)-b*x*mgcos(β)
which looks like SHM except for the constant acceleration term, mgsinβ.
But we note that this makes F, and hence the acceleration, 0 when x = tan(β)/b. So that must be the equilibrium point.
If we measure the distance from x = tan(β)/b instead, i.e. distance = u = x - tan(β)/b, we get F = -b*u*mgcos(β)
Since u is just a constant different from x, the acceleration of x is the same as the acceleration of u, so this is now recognisably SHM, and you can use your standard knowledge about SHM in relation to frequency, amplitude... The only thing to remember is that u is not measuring distance from the top.

 

[squareroot]

Ok, so now i can say that k=-b*mgcos(β) so that period T=2π√(m/k), T=2π√(m/-bmgcos(β)) , T=2π√1/(-bgcos(β)) , and the frequency is 1/T.

But how do i get from this equations to those which help me find the actual distance and time that i need?

 

[haruspex]

Ok, so now i can say that k=-b*mgcos(β) so that period T=2π√(m/k), T=2π√(m/-bmgcos(β)) , T=2π√1/(-bgcos(β)) , and the frequency is 1/T.

But how do i get from this equations to those which help me find the actual distance and time that i need?

The object starts at rest and we want the time until it is at rest again. Whether we measure distance as x or u does not affect the speed, so it's the time between successive points of zero speed in the SHM equation. How does that relate to the period of oscillation?
Similarly, amplitude is half the difference between extremes of position, and for that it won't matter whether we use x or u, so it's the amplitude you see in the SHM equation.