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!
welcome to pf! 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)
Yes to all of that. (For a flexible rope, at least.) By direction of tension I assume you really want the direction of the force due to the tension. Just remember that ropes can't push. So at any point along the rope, each side pulls against the other with a force tangent to the rope.
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
Nothing wrong with your reasoning. We usually (at least in elementary problems) treat the rope as being massless, so there is no net force on any element. (Real ropes have mass and weight, of course.)
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.
Note that the rope is not massless--it has weight W. The net force on the rope--including gravity--will equal zero. That should allow you to figure out the tension in the middle. (If you want the tension in the middle, you should consider one half of the rope.)
Might this belong in the homework section? 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 left-hand tree.
hi david45610! (just got up :zzz:) in equilibrium, the net force must be zero, mustn't it? 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 (you mean "T(middle) to the right" ) 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?