Sliding book on a moving shelf

[NATURE.M]

Homework Statement

A copy of The History of the Decline and Fall of the Roman Empire by Edward Gibbon lies exactly in the middle of an otherwise empty 6.1m long bookshelf in Robarts Library. The coefficient of static friction for the book on the shelf is μS = 11/60 and the shelf has been polished (so kinetic friction may be neglected). A tall, disgruntled librarian slowly lifts up the one end of the bookshelf at a constant rate, so that it pivots upward, anchored at the opposite end (see Figure 2). After 5.5 seconds, she sees the book begin to slide down the
tilted shelf. She continues to raise her end of the shelf at the same rate, gleefully watching until the book falls onto the floor.

How long does it take the book to fall off the shelf (measured from the time it
starts moving)?

Homework Equations

The Attempt at a Solution

So I am pretty sure this problem involves differentiation, since the angle is constantly changing, and which would mean the acceleration of the book is also changing. This relationship is described mathematically as a=gsinθ, where θ is the angle of inclination.
And so the angle is then changing with time, which in turn means the acceleration is changing with time. But how can i relate these quantities in such a way to solve for t?
I tried a number of different approaches (none of which seemed to get me anywhere)
For instance, I took the derivative of t=sqrt(6.1/9.81sinθ) with respect to θ, and tried to use the pythagorean theorem by treating the problem as a related rate, but that didn't get me anywhere.

Oh and sorry for the uninformative title, I realized only after i submitted the post, that I didn't finish the title.

 

[Simon Bridge]

The shelf is lifted at a constant angular velocity $\dot{\theta}$ so there will probably be a differential equation involved... you have a time $t_1$ say when something special happens - it's probably going to turn out to be an "boundary condition" - so this may be an "initial value problem".

You need to draw a free-body diagram for the forces on the book - which will depend on the angle.
You also need an equation for the critical time when $t=t_1$ and what happens then, as well as what happens after.

 

[haruspex]

It's not completely clear whether it's the angular rate that's constant or the rate of vertical movement of the raised end. If the one leads to an intractable equation, try the other.
Please post whatever equations and attempts you've made.

 

[NATURE.M]

The problem shouldn't involve any angular quantities since we haven't started circular motion yet. There should be a simple way to solve the problem with only knowledge of forces, and simple differentiation, but I can't find such a method.

 

[NihalSh]

imagine this to be a situation of an object sliding on a wedge. Then it becomes pretty clear that you have to balance friction force and gravitational force along the wedge at t=5.5s, because at that instant the books just starts sliding. You'll find:

$θ=tan^{-1}μ$...at t=5.5 s

you will find angular velocity to be
$$ω=\frac{θ}{5.5 s}$$

 

[Simon Bridge]

OK - you are the one doing the course: do you understand the question to mean that the angle changes at a constant rate or that the librarian-end of the shelf is lifted with a constant speed?

Either way, my suggestion stands.
Have you attempted the problem yet?

 

[NATURE.M]

I've figured out that the librarian lifts the shelf at a constant rate of 0.20m/s.
And I know the angle and acceleration are related by a=gsinθ. But since this isn't constant acceleration, I'm not sure how to proceed? I understand you could use angular velocity, but since we are still covering dynamics, I'm sure there must be a simpler solution.

 

[NATURE.M]

OK - you are the one doing the course: do you understand the question to mean that the angle changes at a constant rate or that the librarian-end of the shelf is lifted with a constant speed?

Either way, my suggestion stands.
Have you attempted the problem yet?

I interpret it to mean that the librarian-end of the shelf is lifted with a constant speed.

 

[NihalSh]

I interpret it to mean that the librarian-end of the shelf is lifted with a constant speed.

angular speed would be more useful compared to linear speed you have calculated.

you can find velocity by energy conservation, as the question say kinetic friction can be ignored so there is no dissipation of energy.

$$a=\frac{dv}{dt}$$
$$a.dt=dv$$

just find the acceleration as a function of time and it will be a piece of cake. I have already given you the clues: no friction, angular velocity is constant.

Edit: Linear speed is not a constant, it varies. But it can be assumed constant for a short interval.

 

[NihalSh]

I've figured out that the librarian lifts the shelf at a constant rate of 0.20m/s.
And I know the angle and acceleration are related by a=gsinθ. But since this isn't constant acceleration, I'm not sure how to proceed? I understand you could use angular velocity, but since we are still covering dynamics, I'm sure there must be a simpler solution.

I believe the previously given solution would be simpler. But if you still want, here's the alternative:

$$Sinθ=\frac{st}{L}$$
$s$ is the vertical speed with which the edge of shelf is being lifted.
$L$ is the total length of the shelf.

Edit: this equation is only a approximation as speed is not exactly constant. Its better to go with the previously mentioned method.

 

[NATURE.M]

I believe the previously given solution would be simpler. But if you still want, here's the alternative:

$$Sinθ=\frac{st}{L}$$
$s$ is the vertical speed with which the edge of shelf is being lifted.
$L$ is the total length of the shelf.

I obtained this equation as well when trying to solve the problem, but the value of theta is not known since its constantly changing. So,if I use a=gsinθ, I obtain a=(gst/L). I think this would be correct. And so I have acceleration as a function of time.

 

[NihalSh]

I obtained this equation as well when trying to solve the problem, but the value of theta is not known since its constantly changing. So,if I use a=gsinθ, I obtain a=(gst/L). I think this would be correct. And so I have acceleration as a function of time.

Yes, that is what you'll have to use since $sinθ$ is not constant. Integrate with proper limits and you should get the answer.

Important, as haruspex previously mentioned your question hasn't made clear if the height is constantly increased or the angle. But they can both be approximated constant for the given situation (short time period).

 

[NATURE.M]

Yes, that is what you'll have to use since $sinθ$ is not constant. Integrate with proper limits and you should get the answer.

Important, as haruspex previously mentioned your question hasn't made clear if the height is constantly increased or the angle. But they can both be approximated constant for the given situation (short time period).

Okay thanks. I'll try it out and see what I get.

 

[NihalSh]

Okay thanks. I'll try it out and see what I get.

great, keep us posted!

 

[Simon Bridge]

This is HS level right?
Senior level and pre-rotational motion.
Done much calculus?

A simplistic approach:
1. find the angle greater than which the max static friction is overcome by gravity.
2. find the height of the book for this angle
3. use conservation of energy to work out the final speed - in the assumption that further movement of the shelf adds no energy to the book (it still has to fall the same distance)

However.
a) is the shelf being lifted faster than the book falls - at any time?
b) lifting the shelf pushes on the bottom of the book - would this affect the static friction?
c) does lifting the shelf add a force-component along the shelf which could help accelerate the book?

You may have already past that.
But we cannot tell because you don't show us what you have done or how you are thinking about it.

@NihalSh: Isn't handing out equations for consideration sort-of spoonfeeding?
You are getting lucky that OP has already found them ... I've found it is usually better to ask OP for what they have already tried before handing out our own hints.

 

[NATURE.M]

So I've solved the problem integrating the acceleration function to obtain the position function, and just substituting from there. But should we consider L in the equation a=(gst/L) to just simply mean the total length of the shelf (or 6.1m)? I've been thinking maybe we could let L=3.05m instead,
since the book is initially positioned on the middle of the shelf. Then we would still have sinθ=(st)/L, only now L=3.05m. Does this make sense as it pertains to the problem?

 

[NihalSh]

@NihalSh: Isn't handing out equations for consideration sort-of spoonfeeding?
You are getting lucky that OP has already found them ... I've found it is usually better to ask OP for what they have already tried before handing out our own hints.

yes, it is. Sorry about that. I'll keep that in mind next time I go around helping someone!

 

[NihalSh]

So I've solved the problem integrating the acceleration function to obtain the position function, and just substituting from there. But should we consider L in the equation a=(gst/L) to just simply mean the total length of the shelf (or 6.1m)? I've been thinking maybe we could let L=3.05m instead,
since the book is initially positioned on the middle of the shelf. Then we would still have sinθ=(st)/L, only now L=3.05m. Does this make sense as it pertains to the problem?

I guess you are missing the point. we are dealing with $sinθ$ here. $Sin$ function is a ratio of two quantities. If you take different Length, you'll have to adjust the speed accordingly.

Remember what Simon Bridge said, its always assumed (implied) here that book falls faster than the shelf is being lifted. I'll leave to think what would happen if its not the case!

 

[Simon Bridge]

It is important to understand the equations you use - don't take people's word for it... we don't actually do you homework for you, but we can give you hints. We cannot help you if we cannot see how you are thinking about the problem.

i.e. how high has the shelf been lifted before the book starts moving?

If $\theta$ is the angle to the horizontal, the condition is $\mu = \tan\theta$ ... or is it $\mu=\sec\theta$ ?? How would you find out?

Do you know how to relate the trig functions sin, cos, tan, to the dimensions of the shelf?

Considering that the shelf always moves at a constant speed and the book starts out stationary - isn't it certain that there is some period of time where the shelf is lifting the book faster than the book can fall?

The trouble is - I cannot tell if the person setting the problem wants this to be taken into account.

 

[NATURE.M]

I guess you are missing the point. we are dealing with $sinθ$ here. $Sin$ function is a ratio of two quantities. If you take different Length, you'll have to adjust the speed accordingly.

Remember what Simon Bridge said, its always assumed (implied) here that book falls faster than the shelf is being lifted. I'll leave to think what would happen if its not the case!

Okay, and there was one other question that I had. When I performed the calculation, I used t=5.5s as the lower limit of integration, but it gave me a rather odd answer. I obtained 6.1s, which implies the books takes only 0.60s to fall off the shelf. Intuitively speaking, it dosen't make sense for the object to cover 3.05m in 0.60s, with its acceleration, a=gsinθ. So should the lower limit of integration be t=0s?
Sorry for all the questions, it just seems the more I dig into this problem, the more questions that arise.

 

[NihalSh]

Okay, and there was one other question that I had. When I performed the calculation, I used t=5.5s as the lower limit of integration, but it gave me a rather odd answer. I obtained 6.1s, which implies the books takes only 0.60s to fall off the shelf. Intuitively speaking, it dosen't make sense for the object to cover 3.05m in 0.60s, with its acceleration, a=gsinθ. So should the lower limit of integration be t=0s?
Sorry for all the questions, it just seems the more I dig into this problem, the more questions that arise.

The answer should lie between 1 and 2 seconds. That should be obvious looking at the initial acceleration. Check your integration. I did the integration myself using both ways known, and the answer came out to be reasonable and in the range I already gave. Show your work so I can help you figure out the mistake. Your lower limit is correct.

 

[NATURE.M]

The answer should lie between 1 and 2 seconds. That should be obvious looking at the initial acceleration. Check your integration. I did the integration myself using both ways known, and the answer came out to be reasonable and in the range I already gave. Show your work so I can help you figure out the mistake. Your lower limit is correct.

After integrating the acceleration function I have v=(gs)/(2L) * (t^2-ti^2)
where ti=5.5s and s=0.20m/s

Integrating this for position I have,
x=xi+(gs)/(6L)*(t^3-ti^3)
and since x=0m, we have -xi=(gs)/(6L)*(t^3-ti^3), rearranging we have
t=cubicroot((-6Lxi)/(gs) +ti^3)

 

[NihalSh]

After integrating the acceleration function I have v=(gs)/(2L) * (t^2-ti^2)
where ti=5.5s and s=0.20m/s

Integrating this for position I have,
x=xi+(gs)/(6L)*(t^3-ti^3)
and since x=0m, we have -xi=(gs)/(6L)*(t^3-ti^3), rearranging we have
t=cubicroot((-6Lxi)/(gs) +ti^3)

First of all I'll recommend not to go that route!. do not integrate that. Apply energy conservation. The simple reason I can give you for that is you cannot find position at t=0, that is you cannot find $x_{i}$. Also the position function you obtained is incorrect because $t_{i}$ is constant. Moreover, what you'll obtain would be cubic equation of the form $ax^3+bx+c=0$.

I've already mentioned in my previous posts how to obtain velocity $v$.

 

[haruspex]

in the assumption that further movement of the shelf adds no energy to the book (it still has to fall the same distance)

I think that's a wrong assumption. The librarian is doing work lifting the book.
I found this problem so confusing I went back to first principles.
If the slope when movement starts is θ, at some instant it has increased by α, the speed normal to the shelf is u, and the distance from the fixed end of the shelf is x:
$g \sin(\theta+\alpha) = -\ddot x + u\omega$, $u = x \omega$.
Taking ω as constant, α = ω t. From that I get a solution involving trig and hyperbolic trig functions.

 

[NATURE.M]

First of all I'll recommend not to go that route!. do not integrate that. Apply energy conservation. The simple reason I can give you for that is you cannot find position at t=0, that is you cannot find $x_{i}$. Also the position function you obtained is incorrect because $t_{i}$ is constant. Moreover, what you'll obtain would be cubic equation of the form $ax^3+bx+c=0$.

I've already mentioned in my previous posts how to obtain velocity $v$.

At t=0s isn't the position just 3.05m if we let the bottom of end of shelf be the origin.

 

[NihalSh]

At t=0s isn't the position just 3.05m if we let the bottom of end of shelf be the origin.

No. The position of the book at t=5.5 is 3.05.Try to see through the equation. Every equation tells a story. Your velocity equation tells that before t=5.5s, the velocity of book is negative. That means book cannot be at the position you mentioned.

 

[haruspex]

you can find velocity by energy conservation,

I don't understand how. As the librarian continues to raise the shelf, more PE is being added. The rate depends on how far the book has slid.

 

[NihalSh]

I don't understand how. As the librarian continues to raise the shelf, more PE is being added. The rate depends on how far the book has slid.

Its an assumption, we are assuming rate of acceleration is much more than the rate at which potential energy is added. I don't see any other way, to solve this question without making some assumptions. And distance traveled along the shelf is same. The value of initial acceleration clearly indicates time should lie between 1 and 2 seconds. The answer comes out to be reasonable.

 

[Simon Bridge]

The assumption is not valid - since the book starts at rest wrt the shelf, there is a time period when the shelf adds potential energy faster than the book loses it.

Its the sort of thing a postgraduate would maybe consider Lagrangian methods for.

It may work by doing the calculation in the frame of the book ... gravity becomes a force that varies with time, and the component along the shelf is opposed by the centrifugal (pseudo)force. You end up with a second order DE.

Start time at the critical angle that the book starts to move:
i.e. when $\mu mg\cos\theta=mg\sin\theta$ which gives: $\theta(0)=\arctan\mu$

$g\sin(\omega t+\arctan\mu)-x\omega^2=\ddot{x}: x(0)=L/2, \dot{x}(0)=0$

... or something?
This is assuming that the angular velocity is a constant.

The main problem with that is OP lessons have yet to include rotational motion.
A similar method may be possible in the frame of the librarian.
I have a feeling that this is an example of a problem set without thinking it through.

 

[NihalSh]

@Simon Bridge
I know about that technicality, but I was trying to avoid it. You already know this is a basic Physics question. The poster hasn't even studied rotation yet. I believe the question was set with a mindset that the student will apply conservation of energy (although you know why this cannot be applied). If someone has a better method to go about it, I would really appreciate it.

 

[haruspex]

@Simon Bridge
I know about that technicality, but I was trying to avoid it. You already know this is a basic Physics question. The poster hasn't even studied rotation yet. I believe the question was set with a mindset that the student will apply conservation of energy (although you know why this cannot be applied). If someone has a better method to go about it, I would really appreciate it.

Pls take a look at my post #24.

 

[NihalSh]

Pls take a look at my post #24.

yeah, looks like OP would have to solve that differential to get the precise result and would also have to get familiar with rotational variables!

Edit: If we use this method (which we should), the breakaway condition would change:

$$μ.g.Cos(ω.t)+(3.05).ω^2=g.Sin(ω.t)$$

 

[haruspex]

yeah, looks like OP would have to solve that differential to get the precise result and would also have to get familiar with rotational variables!

Edit: If we use this method (which we should), the breakaway condition would change:

$$μ.g.Cos(ω.t)+(3.05).ω^2=g.Sin(ω.t)$$

Good point.