How far up the plank will the box slide before coming to rest?

[benabean]

[SOLVED] Challenging friction problem

Homework Statement

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}

Homework 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

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

<br /> \begin{align*}<br /> v - mg\sin\alpha - f = -ma\\<br /> v - mg\sin\alpha - \mu mg \cos\alpha = -ma\\<br /> mg\sin\alpha + \mu mg \cos\alpha - v = ma\\<br /> mg\sin\alpha + Axmg\cos\alpha - v = ma\\<br /> \end{align*}<br />
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.

Any help gratefully received

thanks, b.

 

[nrqed]

Homework Statement

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}

Homework Equations

\vec{F}_{net} = \Sigma \vec{F} = 0

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

f_s &lt;= \mu_s n

f_k = \mu_k n

w = m g

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

<br /> \begin{align*}<br /> v - mg\sin\alpha - f = -ma\\<br /> v - mg\sin\alpha - \mu mg \cos\alpha = -ma\\<br /> mg\sin\alpha + \mu mg \cos\alpha - v = ma\\<br /> mg\sin\alpha + Axmg\cos\alpha - v = ma\\<br /> \end{align*}<br />
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.

Any help gratefully received

thanks, b.

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.

 

[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

 

[nrqed]

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

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

 

[nrqed]

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

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

 

[benabean]

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

 

[nrqed]

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

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).

 

[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

?

 

[nrqed]

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

?

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.

 

[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.

 

[benabean]

Solution

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}Thanks Patrick

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

Continuing on from above:

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

using the quadratic equation we then get

<br /> 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}<br />

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

 

[nrqed]

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.

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