# Challenging friction problem

1. Apr 15, 2008

### benabean

[SOLVED] Challenging friction problem

1. The problem statement, all variables and given/known data
A worker is shoving boxes up a rough plank inclined at angle $\alpha$ above the horizontal. The plank is partially covered with ice, with more ice near the bottom of the plank than near the top, so that the coefficient of friction increases with distance $x$ along the plank: $\mu = Ax$, where $A$ is a positive constant and the bottom of the plank is at $x = 0$.For this plank, $\mu_k = \mu_s = \mu$. The worker shoves the box up the plank so that is leaves the bottom of the plank moving at speed $v_0$. Show that when the box first comes to rest, it will remain at rest if

$v^2_0 \geq {3g\sin^2\alpha\over A\cos\alpha}$

2. Relevant equations
$$\vec{F}_{net} = \Sigma \vec{F} = 0$$

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

$$f_s <= \mu_s n$$

$$f_k = \mu_k n$$

$$w = m g$$

3. The attempt at a solution

initial speed = $v_0$
final speed = $v$

taking $xy$ axes along and perpendicular to the plank:

$n = mg\cos\alpha$

\begin{align*} v - mg\sin\alpha - f = -ma\\ v - mg\sin\alpha - \mu mg \cos\alpha = -ma\\ mg\sin\alpha + \mu mg \cos\alpha - v = ma\\ mg\sin\alpha + Axmg\cos\alpha - v = ma\\ \end{align*}
I get stuck here now, I don't even think I'm heading in th right direction.
I've tried the problem using the work-energy theorem also but I can't get anything to cancel down to the equation needed.

thanks, b.

2. Apr 15, 2008

### nrqed

To me it sounds like the work energy theorem is absolutely the way to go. The work done by the friction will involve an integral but it will be straightforward. You shoul dbe able to find the final position x when the box comes to rest.

3. Apr 15, 2008

### benabean

Still no luck after having another look. I don't have a clue where the $\sin^2$ comes from, and I can't get the fraction:

$W_\textit{applied} = W_\textit{fric} + W_\textit{grav}$

$\Rightarrow$

$W_\textit{applied} = \int_{v}^{v_0} mv_x dv_x = \int_{0}^{v_0} mv_x dv_x = -\frac{1}{2}mv^2_0$

$W_\textit{fric} = \int -f dx = -\int \mu n dx = -\int_{0}^{x} Axmg\cos\alpha dx = -\frac{Ax^2mg\cos\alpha}{2}$

$W_\textit{grav} = -mgx\sin\alpha$

$\Rightarrow -\frac{1}{2}mv^2_0 = -\frac{Ax^2mg\cos\alpha}{2} -mgx\sin\alpha$

$\frac{1}{2}v^2_0 = \frac{Ax^2g\cos\alpha}{2} + gx\sin\alpha$

$v^2_0 = Ax^2g\cos\alpha + 2gx\sin\alpha$

Can anyone show me were I'm going wrong please

4. Apr 15, 2008

### nrqed

Looks good after a quick glance.

But you are not done. Now you must find x (solve the quadratic). Then you must find what must be the minimum x such that the block does not slide down when it is at rest (that's a separate calculation). Finally, impose that condition on the x you found and get a condition on v_0

5. Apr 15, 2008

### nrqed

By the way, I finished the problem using the result you give above for v_0^2 and got the correct answer, so your expression is correct.

Patrick

6. Apr 15, 2008

### benabean

Does the discriminant part of the quadratic cancel down?? (besides the 4g)
The $\sqrt\cos\alpha$ is whats bothering me.

Last edited: Apr 15, 2008
7. Apr 15, 2008

### nrqed

I am not sure what you mean by "besides the 4g). The discriminant is not zero. Keep only the positive root (since x final must be positive).

8. Apr 15, 2008

### benabean

both terms in the square root are multiplied by 4g.
For the discriminant I have $\sqrt{4g^2\sin^2\alpha -4Agv^2_0 \cos\alpha}$
I can't see any substantial canceling to be done though

???

9. Apr 15, 2008

### nrqed

I get a positive sign for the second term.

No, there will be no simplification at that step. Simply keep going. Find the minimum x so that the block does not slide down. Set it equal to the x you found above. Solve for v_0. It works out, just keep going.

10. Apr 16, 2008

### benabean

Finally I have it! Many thanks Patrick, I'm very grateful for your help.

I'll post the solution up for anyone who is curious.

Cheers, b.

11. Apr 16, 2008

### benabean

Solution
Continuing on from above:

$Ax^2g\cos\alpha + 2gx\sin\alpha - v^2_0 = 0$

using the quadratic equation we then get

$x = \frac{-2g\sin\alpha \pm \sqrt{(2g\sin\alpha)^2 - 4(Ag\cos\alpha)(-v^2_0)}}{2Ag\cos\alpha}$

$x =\frac{\sqrt{4g^2\sin^2\alpha + 4Agv^2_0\cos\alpha} - 2g\sin\alpha}{2Ag\cos\alpha}$

For the box to remain stationary $f \geq w_\textit{parallel}$

$Axmg\cos\alpha \geq mg\sin\alpha$

$Axmg\cos\alpha - mg\sin\alpha \geq 0$

$x \geq \frac{\sin\alpha}{A\cos\alpha}$

Substituting the $x$ from the quadratic equation in we get:

$\frac{\sqrt{4g^2\sin^2\alpha + 4Agv^2_0\cos\alpha} - 2g\sin\alpha}{2Ag\cos\alpha} \geq \frac{\sin\alpha}{A\cos\alpha}$

$\sqrt{4g^2\sin^2\alpha + 4Agv^2_0\cos\alpha} - 2g\sin\alpha \geq 2g\sin\alpha$

$\sqrt{4g^2\sin^2\alpha + 4Agv^2_0\cos\alpha} \geq 4g\sin\alpha$

$4g^2\sin^2\alpha + 4Agv^2_0\cos\alpha \geq 16g^2\sin^2\alpha$

$4Agv^2_0\cos\alpha \geq 12g^2\sin^2\alpha$

$Av^2_0\cos\alpha \geq 3g\sin^2\alpha$

$v^2_0 \geq \frac{3g\sin^2\alpha}{A\cos\alpha}$ Q.E.D

12. Apr 16, 2008

### nrqed

You are very welcome. You did a great job! You did it all by yourself (I just corrected the minus sign under the square root). You did all the work, you just needed some encouragement to keep going. You obviously know your stuff quite well.

Best regards

Patrick