Girl on a swing, find tension and force

[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
$T = T_2 \cos30^o + P \cos 30^o$
because it was used when calculating $T$

Remember, $T = F_c + mg$. 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.

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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)

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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 $T = T_2 \cos 30 + P \cos 30$. 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!