# Force of Tension

1. Sep 6, 2009

### pointintime

1. The problem statement, all variables and given/known data

Tarzan plans to cross a gorge by swinging in an arc from a hanging vine. If his arms are capable of exerting a force of 1400 N on the rope, what is the maximum speed he can tolerate at the lowest point of his swing? His mass is 80 kg and the vine is 4.8 m long?

2. Relevant equations

net force = ma

3. The attempt at a solution

I'm not sure how to deal with the force of tension in this problem

net force radial direction = m a radial direction = Fr = Ft + Farm - Fg

sense I don't know what the force of tension is in this problem I got down to this

v = ((.06 kg^-1 m)Ft - 37 s^-2 m^2)^(2^-1)

is the force of tension equal to what?

2. Sep 6, 2009

### Staff: Mentor

Only consider the forces acting on him. How many forces act on Tarzan?

How does Farm relate to Ft?

3. Sep 6, 2009

### pointintime

oh F arm is equal to Ft ???
what about the force of gravity does that increase the force of tension sense his mass is pulling him downwards and his force of his arm is pulling him upwards?

confused on which force act upwards and which act downwards
tension pulls up
force exerted by arm is up
gravity is down

does Ft = Farm

Fg doesn't have any thing to do with tension in this case?

4. Sep 6, 2009

### Staff: Mentor

Yes! The force that he pulls on the rope equals the force that the rope pulls on him. (Opposite direction, of course.)

Answer my question: What forces act on the man? (Hint: Only two forces act on him.)

5. Sep 6, 2009

### pointintime

gravity and tension how
but home come gravity dosen't also pull the rope down creating tension

6. Sep 6, 2009

### Staff: Mentor

Yes, gravity (down) and tension (up) are the only two forces acting on the man.

Gravity pulls the man down, which ends up creating tension in the rope (since the man pulls on the rope). We treat the rope as massless, thus there's no gravity acting directly on the rope.
Centripetal force is just the name given to the net force in the radial direction when something is centripetally accelerated (like in this example). It's not a separate force--it would not appear on a force diagram. When you apply Newton's 2nd law, then you'll use centripetal acceleration.

7. Sep 6, 2009

### pointintime

so the force of tension is just equal to the tension he exerts on his arm what about the force of gravity acting on the man pulling the man down and as a result pulling the rope down even more

the force of tension in the rope
f arm points upwards
force of gravity pulls the man downards creating more tension in the rope

so does the force of tension in the rope equal the force exerted by arm - the force of gravity acting on the man

8. Sep 6, 2009

### Staff: Mentor

The rope tension pulls up on him with the same force that he pulls down on the rope. (At some speed he will be unable to hold on any longer, since he can only pull with so much force. That's what you're trying to find.)
That will be factored in automatically as you write your equation.

No.

No matter what else happens, the man and rope exert equal forces on each other. That's Newton's 3rd law.

9. Sep 6, 2009

### pointintime

so i got 9.8 s^-1 m

10. Sep 6, 2009

### Staff: Mentor

How did you arrive at this result? Show your equation for Newton's 2nd law and how you solved for speed.

11. Sep 6, 2009

### pointintime

net force = m a = Ft - Fg
net force = m a = 1400 N - (80 kg)(9.8 kg^-1 N)
net force = m a = 1400 N - 784 N
net force = m a = 616 N
r^-1 m v^2 = 616 N
v^2 = m^-1 r 616 N
v = (m^-1 r 616 N)^(2^-1)
v = ( (80 kg) 4.8 m (616 N) )^(2^-1)

v = 6.1 s^-1 m

so on free body diagrams we do not lable the centripetal force because it's not a seperate force but just the name of the net froce?

12. Sep 6, 2009

### Staff: Mentor

Good. (I would write the units as m s^-1 or m/s.)

Exactly.