# A lot of questions about statics

1. Jul 20, 2009

### RoyalCat

Case closed. :)

Well, I've got quite a few questions I couldn't solve. I'd love to be pointed in the right direction, because there are some problems I just don't know how to approach at all.
This is just the first batch, I'll be adding a couple more in a while.

I've attached a link to the relevant diagrams at the bottom of the post. Images removed pending approval.

Question 1
A fly is standing on the bottom of a close jar, positioned on one side of a sensitive balance scale which is initially leveled.
Suddenly, the fly rises up, hovers in place for several seconds, and touches back down.
Which of the following sentences most accurately describes what happens?
A. The scales remain leveled all the time.
B. The side with the jar will go down once the fly rises, stay leveled while the fly hovers in place, and come back up when the fly comes down.
C. The side with the jar will go up once the fly rises, stay leveled while the fly hovers in place, and come back when the fly comes down.
D. The side with the jar will go down all the while the fly is in the air.

The answer according to the book is B.
I am completely lost. Why would it go down if the fly is applying LESS force..? And why would it stay leveled if it's still imbalanced? Bleh, I can't make heads or tails of it.

Question 2
In the system in the diagram, the two masses, corresponding to $$W_1$$ and $$W_2$$ are given.
The coefficient of kinetic friction between the floor and the cart is $$\mu$$
There is no friction between the two masses.
It is given that the cart is traveling with a constant velocity, $$V_1$$ to the right, and that the weight is descending at a constant velocity $$V_2$$

A. Calculate the tension in the string.
B. Calculate the coefficient of friction, $$\mu$$
C. What is the magnitude of the horizontal force which the cart exerts on the weight?
D. Calculate the ratio between the horizontal and vertical components of the weight's velocity, $$\frac{V_{2x}}{V_{2y}}$$
E. How would the above results change if the coefficient of friction between the cart and the weight weren't 0?

My attempt:

The forces acting on the weight are:
Its weight, $$W_2$$
The tension from the string, $$T_$$
The normal force from the cart, $$N_{cw}$$

Since it's moving at a constant velocity, then $$\Sigma \vec F=0$$
Therefore:
$$N_{cw}=0$$ (This is the answer to C, and the book agrees with me on it.)
$$T=W_2$$ (This is the answer to A, and the book agrees with me on it)

The forces acting on the cart are:
Its weight, $$W_1$$
The normal force from the floor, $$N$$
The tension in the string, acting on the right of the cart, where its connected to it, $$T$$
The kinetic friction, $$\mu N$$
The tension in the string again, acting on the pulley on the top of the cart, $$T$$
The normal force from the weight, $$N_{wc}=0$$

Since it's moving at a constant velocity, then $$\Sigma \vec F=0$$
Therefore:
$$N=W_1+T=W_1+W_2$$
$$f_k=\mu(W_1+W_2)$$
$$T=f_k$$

Equating the two expressions for the the tension, I receive:
$$\mu=\frac{W_2}{W_1+W_2}$$
The book's answer is:
$$\mu=\frac{2W_2}{W_1+W_2}$$
Where did I go wrong so far? Should I have counted the tension a third time as well?

For every meter the weight descends, it pulls the string, and the cart, one meter to the right. So the ratio $$\frac{V_{2x}}{V_{2y}}$$ is 1.
The book's answer, however, is 1:2. I can't understand this.

Since the normal force between the weight and the cart is 0, any resulting friction would be 0, therefore, not changing the result by one bit.

Question 3
A uniform rod is placed horizontally at equilibrium on a smooth axis through its center. Two masses, $$m$$ are suspended on the rod as illustrated.
A. Will the rod remain horizontal? If not, which way will it turn?
B. What effect does the the depth at which the left mass is hung have on the condition of the system?
C. Is there any importance to the location of the points $$A$$ and $$B$$ (Given that $$l$$ remains the same)?

The book's answers (I'm sure about the book's answer for A, but the one and C looks like it was an answer to a different question, so I'd take it with a grain of salt.):
A. Yes.
B. None.
C. No.

My answers:
A+C. I got lost with the tension forces acting on A and B, so I took the easy way out.
I just calculated the center of mass, ignoring the various tensions as they are internal forces. It's on the axis of rotation, so the net moment on the fulcrum is 0, as long as the horizontal distance of both the masses is the same.

B. The lever arm for both of the identical forces of gravity acting on the masses is the same, so it doesn't matter.

It's just that it looks too easy, so I'd like some confirmation for the way I reached the answers.

Question 4
The force exerted by a stream of water on a rock on its bottom is proportional to the square of the speed of the current and the effective surface area of the rock according to the relation:
$$F=kAV^2$$

It is given that at a certain point where the speed of the stream is $$V_1$$, rocks are dragged whose maximal weight is $$W_1$$
At another point along the stream, the speed of the water is $$V_2=3V_1$$

A. What is the maximal weight of the stones that are dragged at this point, assuming all the rocks have a similar shape (But different dimensions)
B. How will this result be affected if the rocks are cubic or spherical or of a different shape?

The book refers to an example where "Dragged by the stream" means overturned by it.

The book's answers:
A. The stream can carry rocks proportional to the velocity to the sixth power.
$$W_{max} \propto V^6$$
So $$\frac{W_{2 max}}{W_1}=3^6=729$$
B. The result is unaffected by the shape of the rocks.

Following the example, I looked at a cubic rock.
$$W=\rho a^3$$ where $$a$$ is the length of the side of the cube and $$\rho$$ is its density.
$$W \propto a^3$$

The critical situation is when the normal force is about to leave the bottom of the rock. We'll demand that the net torque around the lower left corner of the rock is 0.
The moment of the contact reaction force is 0 as it acts from the lower left corner.
The moment of the force of gravity is $$\frac{Wa}{2}$$
The moment of the force of the stream, which acts from the center of the side of the cube it's meeting, is $$\frac{Fa}{2}$$

We'll demand that these two opposite moments are equal in magnitude.
The result is:
$$V^2=(\frac{\rho}{k})a$$
$$V^2 \propto a \propto W^{1/3}$$
$$V^6 \propto W$$

Giving us the desired result.

However, this approach breaks down when considering a spherical rock, for which any force not acting through the line of action of gravity produces a net moment about the point of contact. I'd love if it someone could explain why the same result applies here.

Relevant image:

2. Jul 20, 2009

### Staff: Mentor

For 1B, think of a frog jumping instead of the fly, at least for the first and last parts of the answer. Think accelerations...

3. Jul 20, 2009

### RoyalCat

Well, that's a bit out of place considering how this is the first question in a chapter all about statics (The concept of acceleration hasn't even been introduced in the book yet).

Other than that, some other things pop into mind. Even if we do consider the forces the fly enacts on the jar, why would it end up in equilibrium once he starts hovering?

I can't make sense of it going back up after he lands, either. If we go by forces, then wouldn't his landing push the scale down further?
And ignoring that, wouldn't the scales stay in place, since this system would be in neutral equilibrium?

Conceptually this should be an easy question, it's the first question in the chapter! But I just can't get my head around it, I apologize in advance if I end up frustrating you a bit, heh, I've spent a bit of time thinking about it, but I haven't found the correct way of looking at it just yet.

4. Jul 20, 2009

### flatmaster

#2 looks very interesting!! And I thought all the simple academic problems of masses and carts and strings had already been thought up!!

5. Jul 20, 2009

### Staff: Mentor

Well if the balance moves, it's already not a statics problem, right?

And actually, it seems like even answer B is not quite right. The principle I'm thinking of is how the force on the platform is the same whether the fly is standing or hovering. Either way, the weight of the fly is offset by the pushing back up of the platform.

When the fly takes off to move upward, that takes more than its weight in force, and hence the platform is pushed down. But when the fly hovers, it is in equilibrium (weight down, force up), so I would think that the balance should go back to neutral. It doesn't take more than the fly's weight to hover.

6. Jul 20, 2009

### RoyalCat

I agree with the second paragraph, but consider that a tank of water with a rock sitting on its bottom would weigh more than a tank of water with the same rock in it, suspended by a string from above and not touching the bottom.

So taking that into consideration, it seems to me that the weight of the fly only acts on the scale when he is ON the scale.

7. Jul 20, 2009

### Staff: Mentor

The hovering fly is not suspended from above, especially when it hovering just barely off the platform...

8. Jul 20, 2009

### RoyalCat

Wouldn't whatever force it's using to suspend itself suffice to make the analogy valid?
Unless of course we bring aerodynamics into the question, saying that the fly's flapping is increasing the air pressure on the scales, which I highly doubt the question is asking us to do.

9. Jul 20, 2009

### Phrak

Um.. where is the center of mass of the fly-jar system? Too much of a hint?

10. Jul 20, 2009

### Staff: Mentor

Well, that's exactly what holds the fly up while it hovers. Think about a helicopter hovering a meter off the ground. That ground is getting beat up pretty hard, eh?

11. Jul 20, 2009

### RoyalCat

If it weren't for the external forces I'd say it is where it started.
Ignoring the external forces, the center of mass remains at rest, on the same line as the fulcrum, how could the system stay at rest if the right side is raised up, then?

Like I said, I'm just getting mixed up with a whole bunch of things here.

And berkeman, I really doubt that is what the question was aiming at. And it still wouldn't explain why the left scale goes back up once the fly settles down (The equilibrium is neutral, as far as I can tell.)

12. Jul 20, 2009

### Staff: Mentor

In order to land, the fly decreases his lift, and he moves down to meet the platform which is rising because...

Interesting point by Phrak about the center of mass. What makes a scale balance balance, anyway? Guess I need to look that one up...

13. Jul 20, 2009

### RoyalCat

Ah, touche. Since his lift is growing ever smaller.
Though you offer a good explanation, I really doubt this is what the question had in mind. A friend of mine tried to go the COM route too, he got nowhere, though.

14. Jul 20, 2009

### queenofbabes

I'd say the CoM isn't moving. The jar is a closed system. When the fly lands the system must return to it's original state, precisely because CoM hasn't moved.

Here's my take:
As fly lifts off, it pushes off with a force greater than its weight, so the platform moves downwards.

When the fly is hovering, the fly pushes down on the air with its weight, which then pushes down on the bottom of the jar with the fly's weight. So to the jar bottom, it's as if the fly was still resting on its bottom, so it will rise up back to its normal level...

I think here's the catch: if this were happening, the jar bottom would push UP on the air, and the air would push UP on the fly, and the fly would rise up together with the jar bottom! (The height at which the fly hovers, with respect to jar bottom, is unchanged).

Yet the question says the fly is hovering in place! So in that case, the fly must be pushing downwards with a force greater than its weight while hovering, so that it "hovers in place" with respect to the surroundings.

I'm not sure what the book wants you to think, but just by looking at the given choices it's quite evident that it's not all that straightforward....

15. Jul 20, 2009

### Phrak

Whoops! My mistake. I mis-recalled that the fly "suddenly jumped off". It "suddenly rises off".
This could get ambiguous.

16. Jul 21, 2009

### RoyalCat

Well, I reckon I'm content with the air explanation. I'll have to see what my teacher thinks of it, though (He's the one who collected the questions in the book).

I've added a fourth question to my first post.

EDIT:
I've just finished a short e-mail correspondence with my teacher (Who is also the author of the book) and he's helped me corroborate the answers I got here. Thanks a lot everyone. :)

The fly pushes off the scales by exerting a force greater than his own weight, thus, making the left scale descend since that force is affecting it just as well through the air. Then he exerts a force smaller than his own weight when coming down, making the left scale ascend.

For the second question, I needed to count the tension a third time. Same goes for the movement. For every one meter the weight descends, the cart only moves half a meter to the right (It's sort of hard to explain why without hand gestures, heh)

He said I nailed it with the third question.

And as for the fourth, the same reasoning applies to translational movement, which is what I was supposed to be looking at.
Comparing the maximum static friction with the force of the stream quickly yields the same relation, $$V_^2 \propto r$$ where $$r$$ is the length of the side of the cube, or the radius of the sphere, or whatnot. For a non-symmetrical object this does of course become a bit more complex.

Case closed, thanks everyone. :)
I'll also be removing the scans, pending his approval, since he seems a bit reluctant to have them online, not that he's said anything expressed against it.

Last edited: Jul 21, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook