Disclaimer: These two problems are part of my homework. I'd prefer not to get a numerical answer, but rather a push in the right direction.
[quote]
[img]http://www.darkvariant.com/prob48_sidepend.gif[/img]
A pendulum has a length L = 1.34 m, with mass m=0.56 kg at the end. It hangs straight down in a jet plane about to take off as shown by the dotted line in the figure. The jet then accelerates uniformly, and while the plane is accelerating, the equilibrium position of the pendulum shifts to the position shown by the solid line, with D = 0.380 m. At the moment shown, the speed of the plane is 62 m/s. Calculate the magnitude of the plane's acceleration.[/quote]

I'm completely clueless on this one. I know that the period T of the pendulum is independant of the mass, and I could calculate the period but I don't see how that will help.

[quote]
A 65 kg circus performer wants to slide down a rope that will snap if the tension exceeds 529 N. (For you to think about: what happens if the performer hangs stationary on the rope?) At what magnitude of acceleration does the performer just avoid breaking the rope?
[/quote]

Here I figured I'd use F = ma and solve for a; unfortunantly the computer says this answer is wrong.

BTW, the homework is due at 10pm tonight (Mountain Time), but if you want to reply after that time feel free, because I need to know how to do these problems anyway.

First question:

You say you can calculate the period of the pendulum.
Now, note that at the moment shown, exactly 1/4 of a period has passed.
(The mass has gone from amplitude point to equilibrium point.)
In that time, it has gone from 0 to 62 m/sec.
This should be enough clues (hopefully *my* direction is correct, I didn't try to solve it, and I'm not sure about it... Haven't done pendula in a long time.).

Second question:

Forces on the person:
Weight (down)
Rope tension (up)

The person is sliding down, so he has an acceleration 'a' downwards, and the sum of the forces (downwards) is equal to ma.
The rope tension is maximum and the weight and mass are known, solve for a. (Is that what you were already doing?)

Edit: A point to think about in the second question.

Assume another person hangs on the rope without moving, and the rope doesn't snap. Would this person be able to cause it to snap by sliding down? If not, how could he cause it to snap?

Thanks, your explanation of the first problem really cleared things up. However, for the second, I still don't particularly understand. My reasoning went like this:

F = ma. Limiting force is known, mass is known:
a = F / m = 529 / 65.

But this answer is incorrect. Obviously there is another force acting here that I'm missing. Or are you saying:

9.8*65 = W_gravity
65 * a = F_tension - W_gravity
65 * a = 529 - 9.8 * 65

(where a is in the negative direction?)

[quote author=K link=board=36;threadid=5359;start=0#msg45112 date=1077241850]
Obviously there is another force acting here that I'm missing. Or are you saying:

9.8*65 = W_gravity
65 * a = F_tension - W_gravity
65 * a = 529 - 9.8 * 65

(where a is in the negative direction?)
[/quote]
No, there are no other forces. And yes, that's what I'm saying.

The only 2 forces on the person are up (tension - T) and down (weight - W). The sum of the forces is *down* since the person is sliding down at an acceleration a.
So define down as the positive direction and you get:

Sigma F = ma
W - T = ma
mg - T = ma
a = g - T/m

g = 9.8 m/sec^2, T = 529 N (since the rope is "just" about to break, so maximum tension), and m = 65 kg gives an aproximate answer of: a = 1.66 m/sec^2 (positive, so down).

Do it like you did (with the positive direction taken as up) and you'll get the same answer, only negative (because the person is sliding down).

[quote author=K link=board=36;threadid=5359;start=0#msg44978 date=1077158798]
[quote]
[img]http://www.darkvariant.com/prob48_sidepend.gif[/img]
A pendulum has a length L = 1.34 m, with mass m=0.56 kg at the end. It hangs straight down in a jet plane about to take off as shown by the dotted line in the figure. The jet then accelerates uniformly, and while the plane is accelerating, the equilibrium position of the pendulum shifts to the position shown by the solid line, with D = 0.380 m. At the moment shown, the speed of the plane is 62 m/s. Calculate the magnitude of the plane's acceleration.[/quote]
[/quote]

Since you've solved this already, I figured I'd give you my comments. The way I see it, you have a new equilibrium point meaning that the net acceleration affecting the pendulum has the direction of the pendulum. Since you can add accelerations as vectors, I think this means that the magnitude of the acceleration a can be calculated from

a / 0.380 = g / sqrt(1.34**2 - 0.380**2)

a = 2.9 m/s**2