Find the tension T2 of the box hanging from the rope if 1 = 25°, 2 = 65°, T1 = 60.58 newtons, and W = 143.4 newtons
About 130 N
To find the tension T2 of the box hanging from the rope, we can start by drawing a free-body diagram for the box. In the diagram, denote T1 as the tension in the rope connected to the box at an angle of 25° and T2 as the tension in the rope connected to the box at an angle of 65°. The weight of the box, W, acts vertically downwards.
Now, let's break down the forces acting on the box along the vertical and horizontal directions.
In the vertical direction:
- T1 * sin(25°) - T2 * sin(65°) - W = 0
This equation represents the equilibrium condition in the vertical direction. The vertical components of T1 and T2 must balance the weight of the box for it to remain stationary.
In the horizontal direction:
- T1 * cos(25°) + T2 * cos(65°) = 0
This equation represents the equilibrium condition in the horizontal direction. The horizontal components of T1 and T2 must cancel each other out for there to be no net horizontal force.
Now, we can solve these equations to find the values of T1 and T2:
From the vertical direction equation:
T1 * sin(25°) - T2 * sin(65°) - W = 0
Substituting the given values:
T1 * (sin(25°)) - T2 * (sin(65°)) - W = 0
(60.58 N) * (sin(25°)) - T2 * (sin(65°)) - (143.4 N) = 0
From the horizontal direction equation:
T1 * cos(25°) + T2 * cos(65°) = 0
Substituting the given values:
T1 * (cos(25°)) + T2 * (cos(65°)) = 0
(60.58 N) * (cos(25°)) + T2 * (cos(65°)) = 0
Now, you can solve these two equations simultaneously to find the value of T2.