Girl on a swing, find tension and force

[Feodalherren]

Homework Statement

The girl weighs 30kg and her COM is at G. As she is passing the lowest position of the swing with rope length L=4.5m her speed is 3.6m/s. The swing and rope have no mass.

Calculate
a) the tension in the rope
b) the force P that is aligned along the girl's arms

see the picture below

Homework Equations

mechanics

The Attempt at a Solution

This seems too easy...

 

[TomHart]

If she was sitting on a stationary swing, the tension would be equal to the weight. However, this swing has a speed of 3.6 m/s so that equation is no longer true. The greater her speed, the greater the tension. You need something to relate those 2 parameters.

 

[Feodalherren]

Ah of course, the centripetal force! I haven't done any physics since I graduated college and I'm trying to remember all the stuff that I learned years ago :).

If I call the centripetal force Fc= mv^2 / L

Then T= Fc + mg

and

P = T / cos(30)

correct? :)

 

[collinsmark]

Ah of course, the centripetal force! I haven't done any physics since I graduated college and I'm trying to remember all the stuff that I learned years ago :).

If I call the centripetal force Fc= mv^2 / L

Then T= Fc + mg

That looks like a good start.

and

P = T / cos(30)

correct? :)

Be careful on part (b). Examine Figure 8 carefully. The force P -- the force aligned along the girl's arms -- is not the only force with an upwards component.

 

[Feodalherren]

Is the centripetal force also pointing up in case B then? This is where I always got confused with the centripetal force, its direction almost seems arbitrary.

 

[collinsmark]

Is the centripetal force also pointing up in case B then? This is where I always got confused with the centripetal force, it's direction almost seems arbitrary.

Yes, it's directed up; she is accelerating in the up direction.

Regarding the tricky part I mentioned in my last post:

Consider the forces above and below where her hands are on the rope.

Above her hands, the tension on the rope is T. I think we all agree on that.

But what about below her hands? You know that the force P, acting along the girl's arms, contributes to the tension. But is that all? Is there anything else?

 

[Feodalherren]

I think I was thinking about it wrong, I think I was approaching it as a statics problem where the sum of the forces is equal to zero when I should have thought about it as Fc=T-mg. It's starting to make more sense now.

As for forces below her hands... Hmm, would it be mg? That's the only one I can think of.

edit: but that would be down...

I really don't know, I can't think of another force that's pointing up.

 

[collinsmark]

I think I was thinking about it wrong, I think I was approaching it as a statics problem where the sum of the forces is equal to zero when I should have thought about it as Fc=T-mg. It's starting to make more sense now.

As for forces below her hands... Hmm, would it be mg? That's the only one I can think of.

edit: but that would be down...

I really don't know, I can't think of another force that's pointing up.

What about the part of rope below her hands -- the part attached to the base of the swing?

 

[haruspex]

As for forces below her hands

Consider the joint where her hands meet the rope. The only acceleration at the lowest part of the swing is centripetal, which is straight up. So the net force acting on that joint must be straight up. The tension in the rope above the joint acts straight up, but that in her arms is not straight down. What force on that joint completes the picture?

 

[Feodalherren]

There would be another tension force in that rope, but wouldn't it be facing down?

 

[haruspex]

There would be another tension force in that rope, but wouldn't it be facing down?

It would be pulling down on that joint, yes. Why is that a problem? What will it do at the other end of that rope section?

 

[Feodalherren]

Equal but opposite. But collinsmark mentioned something about another force pointing up, I do not see it. And even with another force of tension, I now have too many unknowns to solve the problem. I don't know what the acceleration in x is. I have unknowns T2, P and Fx.

 

[haruspex]

collinsmark mentioned something about another force pointing up

which is what that lower section tension will do at its lower end.

with another force of tension, I now have too many unknowns

You have more unknowns, but also more equations. There is a force balance at the joint with the hands. You can ignore the mass of the hands, so you do not need to consider centripetal force in that one.

 

[Feodalherren]

No I'm completely lost now. This doesn't make any sense to me anymore.

At the point of contact with the hands, T2 would be pointing down, no?

Seen from the perspective of just the swing, it would be pointing up, but would mg even exist since it's already taken up by the girl? And if the swing has no mass, why would there then even be a T2?

 

[collinsmark]

No I'm completely lost now. This doesn't make any sense to me anymore.

At the point of contact with the hands, T2 would be pointing down, no?

Seen from the perspective of just the swing, it would be pointing up, but would mg even exist since it's already taken up by the girl? And if the swing has no mass, why would there then even be a T2?

Maybe it's easier to think of things in terms of tensions.

The vertical component of the tension in her arms together with the vertical component of the tension in the rope below her hands must combine to a total, vertical force T.

Recognizing that, you could draw a free body diagram showing the forces [on the girl].

Does that help?

 

[Feodalherren]

But I have too many unknowns now.

 

[haruspex]

No I'm completely lost now. This doesn't make any sense to me anymore.

At the point of contact with the hands, T2 would be pointing down, no?

Seen from the perspective of just the swing, it would be pointing up, but would mg even exist since it's already taken up by the girl? And if the swing has no mass, why would there then even be a T2?

I think you are confused by the dual role of the arms as both a tension member, whose tension is of interest, and as part of the girl. To clarify matters, replace the arms by another section of rope. So you have one vertical rope at the top joined to two angled ropes at a joint. Those two ropes go to different parts of what remains of the girl (which you can treat as a rigid body and the only mass present).

 

[Feodalherren]

I think you are confused by the dual role of the arms as both a tension member, whose tension is of interest, and as part of the girl. To clarify matters, replace the arms by another section of rope. So you have one vertical rope at the top joined to two angled ropes at a joint. Those two ropes go to different parts of what remains of the girl (which you can treat as a rigid body and the only mass present).

Okay that's helpful, but I still don't see how the T2 helps me at all. The problem says that the swing and rope are massless. So in the FBD of just the lower part of the swing, what is the force pulling it down? What is counteracting T2? How do I find T2?

 

[collinsmark]

I would draw the FBD considering the forces acting on the girl.

What must the vertical components of T₂ and P add to? (Hint: it's not mg. She's accelerating up, remember. But it is something you've already calculated.)

 

[Feodalherren]

that makes a lot more sense. For some reason I was having a really hard time drawing a fbd of the girl.

So then I get

mv^2/L = T - T2cos30 -Pcos30
mg=T2cos30 + Pcos30

correct?

 

[collinsmark]

that makes a lot more sense. For some reason I was having a really hard time drawing a fbd of the girl.

So then I get

mv^2/L = T - T2cos30 -Pcos30
mg=T2cos30 + Pcos30

correct?

No, not quite.

You had the tension T part correct before back in post #3.

The vertical components of P and T₂ don't sum to mg. (She's accelerating don't forget.) They sum to form what? (How do you describe the vertical force in the rope above her hands again?)

 

[Feodalherren]

Okay so all those three forces must equal the centripetal force then?

Fc = T2cos30+Pcos30-mg
Fc=T-T2cos30-Pcos30

 

[collinsmark]

Okay so all those three forces must equal the centripetal force then?

Yes, in this case. Correct. Newton's second law fits in there if you want to be thorough.

Fc = T2cos30+Pcos30-mg
Fc=T-T2cos30-Pcos30

One of those two equations looks correct to me. But I don't think the other one is correct.

 

[Feodalherren]

Then I'm lost again.

I assume that Fc=T2cos30+P2cos30-mg is correct.

My only other guess would then be that T=T2cos30+Pcos30, but that's a guess and it feels wrong. Why wouldn't both points have the same acceleration in the Y-direction?

 

[collinsmark]

I assume that Fc=T2cos30+P2cos30-mg is correct.

Yes, that looks correct to me.

My only other guess would then be that T=T2cos30+Pcos30, but that's a guess and it feels wrong.

That equation is correct!

You know that the total, vertical upward force from the rope is T, right?

You also know that the total, vertical force acting on the girl from the system of arms and ropes is [itex] T_2 \cos 30^o + P \cos 30^o [/itex].

The total upward force (from rope/arm system) acting on the girl is the same as the total upward rope force, right?

So [itex] T_2 \cos 30^o + P \cos 30^o [/itex] is [itex] T [/itex].

They're not just equal. They are the same thing.

Why wouldn't both points have the same acceleration in the Y-direction?

They do.

So what does that tell you, and what can you say about the relationship between the magnitudes of P and T₂? (Hint: symmetry can help you here.)

 

[Feodalherren]

I guess T2 and P have the same magnitude. But I have to confess, I don't understand you solution at all.

If they both have the same acceleration, then how can you just ignore the centripetal acceleration in one equation and include it in the other?

In order to get T=(T2+P)cos30 it seems to me that you first have to do a force balance where you say that the sum of the forces is equal to 0.

 

[collinsmark]

If they both have the same acceleration, then how can you just ignore the centripetal acceleration in one equation and include it in the other?

Newton's second law deals with mass and acceleration. So whenever you are dealing with masses and their corresponding accelerations, you'll have to deal with resultant forces. In this case the resultant force is the centripetal force.

But ropes and arms can be considered mass-less for this problem. We don't need to invoke Newton's second law when analyzing mass-less things. But rules of tension still hold, which is why I suggested treating things in terms of tensions in a previous post.

Using Newton's second law you can calculate T.

Half of T comes from Pcos30. The other half comes from T₂cos30. Together, Pcos30 and T₂cos30 combine to form T.

In order to get T=(T2+P)cos30 it seems to me that you first have to do a force balance where you say that the sum of the forces is equal to 0.

It's really more along the lines of several smaller forces working together to form a larger effective force. We can do this here because we are working with mass-less things that obey rules of tension.

 

[Feodalherren]

I guess I'm still not sure how you can include the centripetal force in one equation, ignore it in the other, but still say that they are both accelerating at the same rate. Shouldn't the resultant force in both cases be the centripetal force then?

 

[haruspex]

I guess I'm still not sure how you can include the centripetal force in one equation, ignore it in the other, but still say that they are both accelerating at the same rate. Shouldn't the resultant force in both cases be the centripetal force then?

Centripetal force is that force required to produce the centripetal acceleration of a mass. No mass, no force.

 

[Feodalherren]

Centripetal force is that force required to produce the centripetal acceleration of a mass. No mass, no force.

This is so confusing :(

 

[haruspex]

This is so confusing :(

At the joint, you have T, T₂ and P. There is no mass in the joint so no gravity and no resultant force. This part is a simple balance of three forces. It does not "know" about the mass of the girl below.

For the girl, you have T₂, P and mg, and the resultant centripetal force.

 

[collinsmark]

I guess I'm still not sure how you can include the centripetal force in one equation, ignore it in the other, but still say that they are both accelerating at the same rate. Shouldn't the resultant force in both cases be the centripetal force then?

Well, the centripetal force is involved in
[itex] T = T_2 \cos30^o + P \cos 30^o [/itex]
because it was used when calculating [itex] T [/itex]

Remember, [itex] T = F_c + mg [/itex]. Here, T is not only the tension in the upper part of the rope, it's also the total vertical tension in the rope/arm system of anything below her hands. All the individual forces in that lower system must sum to to T.

-----

Maybe it will be more clear if we step back and look at a slightly different problem. Here is a problem that uses rocket thrust instead of tensions. I kept some of the variable names the same though for illustrative purposes.

A rocket ship of mass m is taking off near Earth's surface. The rocket has two engines as shown in the figure below. Its acceleration is measured as a and is directly up.

Part i) Find the total effective vertical thrust, F, of the engine cluster in terms of m and a.
Part ii) Find the magnitude of the thrust of the engine P.

Solution:

Part i)
This part is easy. We just invoke Newton's second law and say, ma = F - mg.

or

F = ma + mg.

Part ii)
Since the two engines are configured symmetrically, and there is no horizontal acceleration, we can say that P and T₂ have equal magnitudes.

The total thrust F = Pcos30 + T₂cos30.

and noting the equal magnitudes, we can say

F = 2Pcos30

or

P = F/(2cos30)

Notice that we never had re-invoke Newton's second law in part ii). In other words we didn't have involve ma in that part of the solution anywhere, nor did we for mg for that matter. That's because they're already a part of F.

As our final answer, we can make the final substitution, substituting ma + mg for F, and say

P = m(a + g)/(2cos30)

---------------

Back to our problem with the girl on the swing:

That's the same reason why we don't have to involve the centripetal force again when we say [itex] T = T_2 \cos 30 + P \cos 30 [/itex]. The centripetal force together with the gravitational force are already a part of T.

[Edit: corrected an oversight.]

 

[Feodalherren]

Ahhh! I get it now. You have no idea how much I appreciate that you take the time out of your day to help random strangers online. Thanks a lot y'all!