Question
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).
(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).
Answers
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
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