# Mass on an inclined plane with friction as a function of distance

• klg.amit
In summary, the problem involves a small body sliding down an inclined plane at an angle of \alpha with the horizon, with a coefficient of friction that varies according to \mu = \beta s. The objective is to find the distance traveled until the body comes to a stop and its maximum speed. The force equation for the plane axis is g\sin{\alpha} - \beta s g\cos{\alpha} = a, which simplifies to a differential equation of the form s'' = A - Bs. To solve this, one can use a different method by multiplying both sides of the equation by \frac{ds}{dt} and integrating once. The result is a separable DE that can be solved to find the distance traveled.
klg.amit

## Homework Statement

A small body starts sliding down an inclined plane. The angle between the plane and the horizon is $$\alpha$$. The coefficient of friction depends on the distance $$s$$ that the body travels according to $$\mu = \beta s$$.
Find the distance that the body travels until coming to a stop and its maximum speed.

2. The attempt at a solution

The force equation for the plane axis which I'm pretty sure I got correctly is
$$g\sin{\alpha} - \beta s g\cos{\alpha} = a$$

That is after simplifying and substituting what is required. This is a differential equation which I'm not sure how to solve. First, is there any problem with treating the acceleration as the second derivative of the distance (rather than of the position, which appears the same in this case). Second when I do try to find a homogenous solution I get an imaginary root which I'm not sure is ok. Am I even on the right track ? Or is there something I'm totally missing ?

Welcome to PF!

Welcome to PF!

(have an alpha: α and a beta: β and a mu: µ )
klg.amit said:
A small body starts sliding down an inclined plane. The angle between the plane and the horizon is $$\alpha$$. The coefficient of friction depends on the distance $$s$$ that the body travels according to $$\mu = \beta s$$.
Find the distance that the body travels until coming to a stop and its maximum speed.

The force equation for the plane axis which I'm pretty sure I got correctly is
$$g\sin{\alpha} - \beta s g\cos{\alpha} = a$$

Yes, essentially you have s'' = A - Bs,

and if you put r = s - A/B, r'' = s'', that's r'' = -Br, which is simple harmonic motion, r = C1cos(√B)t + C2sin(√B)t (except, of course, that it gets stuck when it reaches maximum "amplitude" )
… I get an imaginary root which I'm not sure is ok …

Yes, that's normal for simple harmonic motion … imaginary roots correspond to cos and sin, while real roots correspond to exp.

O.K. Thank you :)

This clears up things a bit I need to look into that and harmonic motion more carefully.. (it is something which I actually haven't studied yet)

I got a clue from somewhere which seems to hint that I can do it another way
The clue says to multiply both sides of the equation by $$\frac{ds}{dt}$$ but I have no clue what this clue gives me :). That is, how to continue from there..

You have given the equation of acceleration.
g*sinα - β*s*g*cosα = a.------(1)
Now a = dv/dt = dv/ds*ds/dt = (v*dv)*(1/ds)
Put this value in eq(1) and simplify. You get
g*sinα*ds - β*g*cosα*s*ds = v*dv.
Find the integration on the both side and put v = 0 to find s.

klg.amit said:
I got a clue from somewhere which seems to hint that I can do it another way
The clue says to multiply both sides of the equation by $$\frac{ds}{dt}$$ …

Yes, if s'' = f(s), you can multiply both sides by s', giving:

s's'' = f(s)s'

The LHS is the derivative of 1/2 (s')2 (incidentally, that's exactly where the 1/2 mv2 in KE comes from ), and RHS is the derivative of ∫f(s)ds.

So the point of this method is to move from a second order (and non-separable?) DE to a separable DE ?

klg.amit said:
So the point of this method is to move from a second order (and non-separable?) DE to a separable DE ?

No , the point is that you start with a second-order equation which needs integrating twice, and first you integrate it once.

How do I integrate this part: β*g*cosα*s*ds
I need to find an expression which when differentiated with respect to s gives s
I thought it might be (β*g*cosα*s^2)/2 but then according to the chain rule when differentiating it should be multiplied by s' . I'm sorry but I'm really weak with DE ..

klg.amit said:
How do I integrate this part: β*g*cosα*s*ds
I need to find an expression which when differentiated with respect to s gives s
I thought it might be (β*g*cosα*s^2)/2 but then according to the chain rule when differentiating it should be multiplied by s' . I'm sorry but I'm really weak with DE ..

Sorry, but you're not making sense

when you differentiate wrt s, there's no chain, is there?

True. In that case I was just weak with making sense. Thank you :)

## 1. What is an inclined plane with friction?

An inclined plane with friction is a simple machine that consists of a flat surface that is tilted at an angle, also known as an incline, and a force that opposes motion called friction. This setup allows objects to be moved up or down the incline with less force than if the object were to be lifted straight up.

## 2. How does the mass affect the motion on an inclined plane with friction?

The mass of an object affects its motion on an inclined plane with friction by increasing the amount of force required to move the object up the incline. The greater the mass, the greater the force needed to overcome friction and move the object up the incline.

## 3. How does the angle of the incline affect the friction on an object?

The angle of the incline affects the friction on an object by increasing the force of friction as the angle increases. This is because the steeper the incline, the more the weight of the object acts perpendicular to the surface, increasing the normal force and thus increasing the force of friction.

## 4. How does distance affect the force of friction on an inclined plane?

The distance on an inclined plane affects the force of friction by increasing the total amount of work required to move an object up the incline. As the distance increases, so does the amount of energy needed to overcome friction and move the object up the incline.

## 5. What is the relationship between the coefficient of friction and the force of friction on an inclined plane?

The coefficient of friction and the force of friction on an inclined plane have a direct relationship. The coefficient of friction is a constant value that represents the amount of friction between two surfaces, and the force of friction is equal to the coefficient of friction multiplied by the normal force. As the coefficient of friction increases, so does the force of friction, making it more difficult to move an object up the incline.

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