# Motion on an inclined plane against a resisting force

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1. Aug 10, 2015

### FallenLeibniz

1. The problem statement, all variables and given/known data
I am trying to prove what is asked in the following problem:

A particle of mass m slides down an inclined plane under
the influence of gravity. If the motion is resisted by a
$f=kmv^2$, show that the time required to move a distance
d after starting from rest is:

$t=\frac{arccosh(e^{kd})}{(kg*sin(\theta))^{\frac{1}{2}}}$

2. Relevant equations

Newton's second law?

3. The attempt at a solution

I begin by drawing an inclined plane and I set my reference
frame axes such that the positive x-axis is pointing up the
plane. This gives the acceleration down the plane as
$kv^2-gsin(\theta)$ via use of Newton's Second Law.
I proceed by writing the acceleration as equal to the
time derivative of the velocity and set up the problem to
be evaluated by seperation of variables:

$\int{\frac{dv}{v^2-\frac{g\sin(\theta)}{k}}}=k\int{dt}$

I had managed to find this form in a table of integrals, but
the form gives three different solutions (given in the
attachment). Now I proceeded by using the
natural log solution, solving for v, then trying to
integrate v with respect to t to get the displacement and
finish with solving for t. The problem is the end result is
that I get a coth expression when I solve for v which ends
up giving $t=\frac{arcsinh(e^{kd})}{(kg*sin(\theta))^{\frac{1}{2}}}$
instead of $t=\frac{arccosh(e^{kd})}{(kg*sin(\theta))^{\frac{1}{2}}}$.
By looking over the math I found that the blood clot is that
there is an implicit assumption in trying to outright solve
for v from the ln expression rather than assuming that the
tanh solution to the integral is the actual one that v is
positive. Can anyone tell me how this comes to occur? Also,
can anyone tell me if they get an arccosh with my approach
rather than an arcsinh?

#### Attached Files:

• ###### WP_20150810_001.jpg
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2. Aug 10, 2015

### Student100

Can you also post or attach your complete work?

3. Aug 11, 2015

### andrewkirk

You have switched the signs mid-problem, because you have distance x pointing up the plane but your formula is for acceleration down the plane. It's best to keep directions consistent, in which case your integral equation for $\ddot{x}$, which is up the plane, is:

$$\int_{v_0}^{v_1}\frac{dv}{a^2-v^2}=k\int_0^t du$$

where $a\equiv\sqrt{\frac{g\, \sin\theta}{k}}$.

Wolfram Alpha offers only a single solution for the left hand integral, which is

$$\left[\frac{\tanh^{-1}\frac{v}{a}}{a}\right]_{v_0}^{v_1}$$

This can be simplified using the fact that $v_0=0$, and it's fairly easy to proceed from there.

4. Aug 11, 2015

### FallenLeibniz

@andrewkirk I am not sure what you mean by "you have distance x pointing up the plane". I did say that only my notion of what a "positive distance" was is going up the incline. Can you elaborate?

@Student100 : I am attaching my work on the problem here...

#### Attached Files:

• ###### WP_20150811_005.jpg
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Last edited: Aug 11, 2015
5. Aug 11, 2015

### andrewkirk

I'm referring to this sentence you wrote in the OP:
"such that the positive x-axis is pointing up the plane"
Then later on you say
" This gives the acceleration down the plane as kv2−gsin(θ)"
However the formula you have given in the second case is for acceleration up the plane, not down, so it is only the words that are wrong, not the formula. Because you have taken the positive direction as up, whereas I have taken it as down, your formula differs from mine by a sign. That is, you will get:

$$\int_{v_0}^{v_1}\frac{dv}{a^2-v^2}=-k\int_0^t du$$
and that's what you have got, so up to there your work is correct. You just need to bear in mind that with your chosen direction the velocity $v$ will always be negative or zero.

Have you tried following the steps I outlined from that point on?

6. Aug 11, 2015

### Student100

I can't really see your work all that well, your approach still seems off to me:

$m \ddot x \vec{x} = −kmv^2\vec{x} + r\vec{y} + mg(\vec{x}sin(\theta)-\vec{y}cos(\theta))$
$\dot x = v$
$m \ddot x = −km \dot x^2 + mgsin(\theta))$
$\ddot x = -k \dot x^2+gsin(\theta)$
...

Actually I just put it in word so I could blow it up, your set up should be fine up until Andrews statement. I misread you earlier and assumed you picked down the slope to mean the positive x direction.

Last edited: Aug 11, 2015
7. Aug 12, 2015

### FallenLeibniz

I was under the impression in my work that the resistive force would never be greater in magnitude than the effect of gravity. In my setup, $$k{v}^2$$ is the resistive acceleration and is pointing up the plane (in the positive x-direction) and $$g\sin(\theta)$$ points down the plane (in the negative x-direction). That would mean that $$k{v}^2-g\sin(\theta)$$ is the acceleration of the mass in general. My use of the phrase "down the plane" only carries the implicit assumption that block will never move up (i.e. that the resistive force will never exceed the effect of the weight on the block's motion). Is there something wrong in that reasoning? I am not trying to be dense, but I would really like to make sure I have the Physics right here before I put this problem to bed.

Last edited: Aug 12, 2015
8. Aug 12, 2015

### FallenLeibniz

@andrewkirk : Figured it out. I kept looking at something wrong, but have seen the blood clot.