How Do You Calculate Car Acceleration from a Hanging Object's Angle?

[AznBoi]

Homework Statement

A 3kg object hangs at one end of a rope that is attached to a the ceiling of a railroad car. When the car accelerates to the right, the rope makes an angle of 4 degrees with the vertical. Find the acceleration of the car.

Homework Equations

(sigma)F=ma ..and I think that's it.

The Attempt at a Solution

First, as always, I drew a free-body diagram of what is described in the problem. When the object is 4 degrees with the vertical, the weight force vector of it is pointed straight down, the tension vector pointed to the top right along the string that attaches it to the ceiling of the car. What about the (x) vectors? Are there two equal magnitude vectors pointing in opposite directions? I'm confused at wether or not the object is still or if it is still moving. I suppose that it is still though.

This is a review problem by the way, and I have lost my acumen for Newtons laws, force diagrams, etc. Please guide me through the steps of solving this problem. By all means, feel free to admonish me on anything that I may be doing wrong. Thanks!

 

[Saketh]

Draw the free body diagram, write out Newton's second law with all the forces (gravity and tension), then solve for the acceleration of the car. You can think of the $\Sigma \textbf{F}$ as a "force" due to inertia.

Here's another way to think about it:

First of all, the object is still with respect to the car. This is because, as you will see, the forces cancel out.

When the car accelerates, a fictional force is created inside the car, and the car is no longer an inertial reference frame. This is very interesting - it is the idea that Einstein had, that a person inside the car won't be able to tell if it's gravity pulling them or if the car is accelerating.

If you are in a car, and the car accelerates really fast, are you pushed back or forward? (Think also about what happens when you decelerate rapidly.)

Now that we have established the direction of this fictional force, we need to find its magnitude. Draw your free body diagram (which you have done correctly) and draw in the fictional force.

So we have three forces on the object - gravity, tension, and the fictional force. When you draw the diagram, you will find that the tension keeps the fictional force in check in the x-direction.

You can find the tension in terms of the gravity, then the fictional force in terms of the tension. Because the fictional force is caused by the car's acceleration, the rest is just straightforward Newton's 2nd law.

 

[f(x)]

Since the f.o.r is accelerated, simply use a pseudo force in the opposite direction in your FBD of the mass and that should be enough
As for the whether the particle(weight) is moving or not...i would say no it isnt. It should be in equilibrium at 4 degrees to the vertical...thats why you can equate all the force acting on it

 

[AznBoi]

So why would you call the force pointing to the left a friction force? I don't see how. I mean the object is 4 degrees to the left because of inertia and the car's acceleration rightward. What does friction have to do with it? Is there another more competent way of naming the force vector?

EDIT: Thanks to both of you by the way! =]

 

[AznBoi]

Okay, I found the frictional force( or w/e). However, how do you find the acceleration? I know that the acceleration causes the frictional force. So obviously they oppose each other. How do I find the acceleration of the car? I know that the object is stationary. It is also, however, relative to the car's motion. hmm interesting. Can anyone explain this to me? Thanks!

 

[Doc Al]

So why would you call the force pointing to the left a friction force? I don't see how.

Saketh was talking about fictional forces, not frictional force! (A better term would be inertial force or psuedo-force.) Such "forces" are artifacts of describing things from a non-inertial frame, such as the accelerating train. My advice is to forget about fictional forces (until you get to them in more advanced courses) and view things from an inertial frame--the train tracks, for example--in which the usual Newton's laws apply without the need to invent any new forces.

Do this:
(1) Identify the forces acting on the hanging object. (Hint: Two forces act on it.)
(2) Analyze the vertical and horizontal components and apply Newton's 2nd law to each. (Hint: In which direction is the object accelerating?)
(3) Solve these two equations together to find the acceleration.

 

[turdferguson]

The point of the FBD is to get rid of fictitous forces. Theres only 2 forces, and only one force that affects the acceleration

 

[AznBoi]

Saketh was talking about fictional forces, not frictional force! (A better term would be inertial force or psuedo-force.) Such "forces" are artifacts of describing things from a non-inertial frame, such as the accelerating train. My advice is to forget about fictional forces (until you get to them in more advanced courses) and view things from an inertial frame--the train tracks, for example--in which the usual Newton's laws apply without the need to invent any new forces.

Do this:
(1) Identify the forces acting on the hanging object. (Hint: Two forces act on it.)
(2) Analyze the vertical and horizontal components and apply Newton's 2nd law to each. (Hint: In which direction is the object accelerating?)
(3) Solve these two equations together to find the acceleration.

Oh, lol. I don't seem to recall fictional forces or inertial frames, even though my teacher might have described it before.

1) The two forces that act on the object would be tension and gravityy?
2) I divided the Tension into x and y components. The object is accelerating to the right I suppose because it is moving relative to the car which is accelerating to the right.
3) I don't get how you can find the acceleration of the car if you only have forces acting on the object? F=ma, but a=0 for the object ?? but still the object is moving! relative to the car, so I don't know how this works.

 

[Dorothy Weglend]

Oh, lol. I don't seem to recall fictional forces or inertial frames, even though my teacher might have described it before.

1) The two forces that act on the object would be tension and gravityy?
2) I divided the Tension into x and y components. The object is accelerating to the right I suppose because it is moving relative to the car which is accelerating to the right.
3) I don't get how you can find the acceleration of the car if you only have forces acting on the object? F=ma, but a=0 for the object ?? but still the object is moving! relative to the car, so I don't know how this works.

The car is accelerating to the right of the object. The object 'wants' to stay where it is, because of inertia, but the rope drags it along. The car must continue to accelerate in order to 'stay ahead' of the object. If the car stops accelerating, then the object will (eventually, after swinging back and forth) hang straight down once again.

Dorothy

 

[AznBoi]

The car is accelerating to the right of the object. The object 'wants' to stay where it is, because of inertia, but the rope drags it along. The car must continue to accelerate in order to 'stay ahead' of the object. If the car stops accelerating, then the object will (eventually, after swinging back and forth) hang straight down once again.

Dorothy

So would the acceleration be the x component of tension?? o.0 Thats in netwons though, so should I divide it by the objects mass?

 

[Doc Al]

Oh, lol. I don't seem to recall fictional forces or inertial frames, even though my teacher might have described it before.

Right--forget about it!

1) The two forces that act on the object would be tension and gravityy?

Right.

2) I divided the Tension into x and y components. The object is accelerating to the right I suppose because it is moving relative to the car which is accelerating to the right.

The object is at rest with respect to the car, so its acceleration equals the acceleration of the car.

3) I don't get how you can find the acceleration of the car if you only have forces acting on the object? F=ma, but a=0 for the object ?? but still the object is moving! relative to the car, so I don't know how this works.

The object is tied to the ceiling of the car; it's not going anywhere relative to the car--but it sure is accelerating (along with the car) relative to the tracks.

The only force accelerating the hanging object is the rope tension--but that tension must also support the weight of the object. That's why the rope hangs at an angle. (As Dorothy explained, if the car stops accelerating the rope will end up vertical.)

 

[Saketh]

I'm sorry if I confused you with the "fictional" forces thing - it was another way of approaching the problem.

So would the acceleration be the x component of tension?? o.0 Thats in netwons though, so should I divide it by the objects mass?

Okay, from your question it seems like you need some more help . Let's start from the beginning.

1. Draw your free body diagram, find the forces, etc. This you have already done. Think of yourself as someone outside of the car, looking at the car.
2. Write down Newton's Second Law in the x-direction and the y-direction.
3. We already know the ball is accelerating (because it is attached to the car, and the car is accelerating according to our eyes) - now it's a matter of how much it is accelerating. That's what Newton's Second Law is for!

So we have Newton's Second Law, which says that an object accelerates as much as the net force on it over the mass of the object.
$$\Sigma F_x = ma$$

So would the acceleration be the x component of tension?? o.0 Thats in netwons though, so should I divide it by the objects mass?

No - the net force would be the x-component of tension, then you would divide by the mass. Go back and refresh your memory on Newton's Law.

Don't blindly plug into this formula - understand what it is that you are doing. Try to do another problem of this type by yourself, and if you do ot correctly you know you're on the right track. Explain to yourself each step of the process; convince yourself that you understand why the ball can hang at an angle like this. I'm not censuring you - I just don't want you to make the same mistakes I did.

 

[AznBoi]

Oooh, I think this is all making sense to me, even the fictitious force you were talking about. Please correct me if I'm wrong.


So on the FBD, you only draw 2 forces acting on the object (tension and gravity), or 3 if you divide tension into x and y components.

I think what I did before was a mistake: I drew 4 forces, y tension, x tension, weight, and another force (pointed to the left). <<-- Is the "another force" the fictitious force?? If so, do you need to draw it? The fictitious force is caused by the acceleration, and the net force (pointed to the right) is opposing the fictitious force to keep the object stationary at 4 degrees with the vertical correct??

I guess you wouldn't draw the ficitious force, you would just assume it. Because if you drew it then there would be no net force of the object. There would be two exact forces opposing each other and the acceleration would be 0. However, the fictitious force is caused my the car so you leave it out of the FBD??


Waiiit. It is confusing me again. So the netforce divided by the mass would give you the acceleration right? Does the fictitious force have the same magnitude as the x component of the tension force? Thanks a bunch!

 

[Dorothy Weglend]

Ooohhh.. you are disregarding the Zeroth law of physics...

But if you must... It's a question of frames. If you are outside the car, then it is obvious what is causing the angle: it is the acceleration of the car away from the object, so in inertial frame of the earth, there is an acceleration, or, in other words, Fnet = ma.

But if you are inside the car, you have an equilibrium condition that you can't explain, since you don't know you are accelerating. This object should be hanging straight down, but instead, it is aiming for the back of the car. It appears to be in equilibrium, since you can't detect an acceleration, but it is aiming to the back all the same. So there must be some 'fictitious' 'mystery' or 'pseudo' force that is pushing it there. In this case, since it is equilibrium:

Fnet - ma = 0

It's the same set of forces, looked at in two different frames.

Dorothy

 

[turdferguson]

My teacher explains it like this: our brains are so in tune with the laws of inertia that in a non-inertial reference frame, we need to think, "Somethings not right, there has to be a force pushing the mass back." This phenomenon is a fictitous force. But under closer examination from another reference frame, its clear that tension causes the acceleration. You should know what a fictitous force is, but you shouldn't deal with them mathematically in physics problems because they arent real.

Something else you should note is that the masses cancel out in your equation. The 3kg is extra information, and the acceleration only depends on the angle of 4 degrees

 

[AznBoi]

My teacher explains it like this: our brains are so in tune with the laws of inertia that in a non-inertial reference frame, we need to think, "Somethings not right, there has to be a force pushing the mass back." This phenomenon is a fictitous force. But under closer examination from another reference frame, its clear that tension causes the acceleration. You should know what a fictitous force is, but you shouldn't deal with them mathematically in physics problems because they arent real.

Something else you should note is that the masses cancel out in your equation. The 3kg is extra information, and the acceleration only depends on the angle of 4 degrees

Wait, don't you need to divide the net force by the mass in F=ma, to get the acceleration? Why is the 3kg extra information?

 

[turdferguson]

Did you do the problem? Fx = max. This means the acceleration in the x equals the x component of tension divided by mass. But you also know that the vertical compontent of tension cancels out weight. tension*cos(theta) = mg. You can then solve for tension plug this into the first equation, and the answer is independent of mass like I said