Register to reply 
Tension Direction? 
Share this thread: 
#1
May2312, 06:11 PM

P: 13

Is the direction of tension always tangent to the rope or thing for which it is acting? For example, lets say that there is a rope hanging with both ends on two trees at equal heights. THen the rope would form a parabolic shape. Would the tension, say, at the middle, then be directed horizontally?
Another question. How do you determine the direction of tension? In the above example, in which direction would the tension in the middle of the rope point? What about at the two ends? Thanks! 


#2
May2312, 06:19 PM

Sci Advisor
HW Helper
Thanks
P: 26,160

hi david456103! welcome to pf!
yes, the tension in a rope at any point is always parallel to the rope at that point (ie the tangent) (and each tiny element of the rope has the tension pulling it at both ends in opposite directions) 


#3
May2312, 06:21 PM

Mentor
P: 41,306




#4
May2312, 06:24 PM

P: 13

Tension Direction?
thanks for the answers tim and Doc Al, it makes more sense now
but if the tension force pulls both ends of a tiny element in opposite directions, wouldn't the net force on each tiny element be 0, and thus the net force on the rope be 0? please correct me if my reasoning is flawed 


#5
May2312, 06:29 PM

Mentor
P: 41,306




#6
May2312, 06:40 PM

P: 13

The reason I asked about tension is because I'm trying to solve the following problem(Kleppner 2.22):
"A uniform rope of weight W hangs between two trees. The ends of the rope are the same height, and they each make angle thetha with the trees. Find: a. The tension at either end of the rope b. The tension in the middle of the rope" I have no problem with part a; the tension at each end points outward(northwest) at angle theta to the horizontal, and from there it is just algebra. I have more trouble with part b. On various online sources, they told me to consider "one half of the rope". Let's say we consider the left half. Then there is a horizontal force T(end)cos(theta) to the left, and a horizontal force T(middle) to the left, so the net force on the left portion wouldn't be zero. 


#7
May2312, 06:49 PM

Mentor
P: 41,306




#8
May2312, 06:52 PM

Sci Advisor
Thanks
P: 3,455

In any case, T(middle) acting on the left half is a force to the right  it's what's keeping the left half from swinging back against the lefthand tree. 


#9
May2412, 01:36 AM

Sci Advisor
HW Helper
Thanks
P: 26,160

hi david45610!
(just got up ) for a massless rope, this proves the tension must be the same everywhere for a massive rope, it usually won't be, eg if it's hanging vertically, then each element of length dz will have T(z+dz) = T(z) + (mg/L)dz, ie dT/dz = mg/L the net force must be zero (in equilibrium) … the two Ts are different, so use that equation, and a vertical equation (including W/2) to find the two Ts … what do you get? 


Register to reply 
Related Discussions  
Surface tension's direction  General Physics  2  
SHM spring system, Tension direction confusion.  Introductory Physics Homework  2  
Find tension and Direction  Introductory Physics Homework  3  
Tension and Friction [Tension in a horizontal direction] Help.  Introductory Physics Homework  12  
Direction of Tension Forces  Introductory Physics Homework  2 