Tension at the ends of the rope.
We believe the rope as the point mass suspended by two massless strings that make the angles ϑ with respect the trees. => there are 3 forces: mg, T₁ and T₂ . T₁=T₂=T. The system is in equilibrium and mg is in y-direction =>
mg-T₁(y)-T₂(y) = 0
mg – 2Tcosϑ =0 =>
T=mg/2cosϑ
In the center of the rope, the tension has only horizontal component. Since the sum of all forces is zero, the x-component at the end of the rope should be cancelled by an equivalent tension at the center of the rope. => The tension at the center will be the x-component of the tension at the end of the rope
T(x) =Tsinϑ= mgsinϑ /2cosϑ=mgtanϑ/2
Suppose a rope of mass m hangs between two trees. The ends of the rope are at the same height and they make an angle θ with the trees.
(a) What is the tension at the ends of the rope where it is connected to the trees? Express your answer in terms of m, g, and θ (enter theta for θ).
(b) What is the tension in the rope at a point midway between the trees? Express your answer in terms of m, g, and θ (enter theta for θ).
1 answer
1 answer