Question

A gymnast of mass 65.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81 m/s^2 for the acceleration of gravity. (a)Calculate the tension (T) in the rope if the gymnast hangs motionless on the rope. (b)Calculate the tension (T) in the rope if the gymnast climbs the rope at a constant rate. (c)Calculate the tension in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.50 m/s^2. Our teacher never went over tension force. Idk how to answer these plz help

Answer 1

For (a), if the figure skater is motionless there is zero acceleration, which indicates that forces are balanced. There are two forces acting on the skater, and must be equal in magnitude and acting in opposite directions to result in a zero net force. See if you can solve this before moving on.

Answer 2

And by figure skater I mean gymnast : - ).

Answer 3

Oh! So I convert 65.0 kg to newtons but multiplying 65 kg with 9.81 m/s^2. I ended up getting the answer that it gave!

Answer 4

Yes, the force of gravity is equal to mass times acceleration. This points in the negative y-direction (if you set up your diagram like that), and the only thing to stop her from accelerating downward is a force pulling her up, which is the tension.

Now for (b) the gymnast is moving at a constant velocity. What does this indicate about the acceleration, and thus, the force?

Answer 5

According to Newtons first law, if an object is in constant velocity has no forces acting on it...right? :O

Answer 6

I mean, forces will act on that object, but it can still have NO net force, right?

Answer 7

Right, so has anything changed?

Answer 8

No So would the answer be zero??

Answer 9

No, it would mean that the net force is zero. She hasn't changed in mass, has she? So the force acting on her due to gravity is the same. Therefore, the tension should also be the same.

Answer 10

Oh, pffft duh! I see it now. So since she is not accelerating, it does not affect the mass in Newtons and I would get the same answer as part A. But now he acceleration changes and is an "upward" for part C. :O Does that mean I would be doing a formula for y-direction? o.O

Answer 11

Yes. Because there is an acceleration the net force is not zero. The net force = \[F_n = F_t - F_g\]

You can find the net force by: \[F = ma\]

See if you can solve for the tension.

Answer 12

I got: Ft= (-9.8m/s^2) / (65kg x 1.50 m/s^2). But that does not seem right.

Answer 13

Well, try and think of the mechanics occurring. The gymnast is pulling on the rope hard enough to make her accelerate at 1.50 m/s^2. This must mean that the tension in the rope is now greater than it was when she was moving with constant velocity, as in part (b).

Remembering that the force of gravity hasn't changed, how would you rearrange the first equation from my last post to find the force of tension.

Answer 14

Ft= (Fg/Fn) ? Fg is not supposed to be negative, sine it is a downward force? I am horrible at physics so I do not mean to make the problem harder than it it is. D: The website is telling me to use Fn=T-mg. Even when I rearrange the equation to get T by itself I still get the answer wrong