# Force of Tension

by pointintime
Tags: force, tension
 P: 166 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?
Mentor
P: 41,323
 Quote by pointintime net force radial direction = m a radial direction = Fr = Ft + Farm - Fg
Only consider the forces acting on him. How many forces act on Tarzan?

How does Farm relate to Ft?
 P: 166 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 radial force is upwards does Ft = Farm Fg doesn't have any thing to do with tension in this case?
Mentor
P: 41,323
Force of Tension

 Quote by pointintime oh F arm is equal to Ft ???
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.)
 P: 166 gravity and tension how but home come gravity dosen't also pull the rope down creating tension and what about centripetal force?
Mentor
P: 41,323
 Quote by pointintime gravity and tension how
Yes, gravity (down) and tension (up) are the only two forces acting on the man.

 but home come gravity dosen't also pull the rope down creating tension
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.
 P: 166 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
Mentor
P: 41,323
 Quote by pointintime so the force of tension is just equal to the tension he exerts on his arm
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.)
 what about the force of gravity acting on the man pulling the man down and as a result pulling the rope down even more
That will be factored in automatically as you write your equation.

 so does the force of tension in the rope equal the force exerted by arm - the force of gravity acting on the man
No.

No matter what else happens, the man and rope exert equal forces on each other. That's Newton's 3rd law.
 P: 166 so i got 9.8 s^-1 m
Mentor
P: 41,323
 Quote by pointintime so i got 9.8 s^-1 m
How did you arrive at this result? Show your equation for Newton's 2nd law and how you solved for speed.
 P: 166 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?
Mentor
P: 41,323
 Quote by 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
Good. (I would write the units as m s^-1 or m/s.)

 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?
Exactly.

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