A 5 000-N weight is held suspended in equilibrium by two cables. Cable 1 applies a horizontal
force to the right of the object and has a tension, T1. Cable 2 applies a force upward and to the
left at an angle of 37.0° to the negative x axis and has a tension, T2. What is the tension, T1?
6640
To find the tension, T1, we need to analyze the forces acting on the object. Let's break down the given information and solve step-by-step.
1. Identify the forces:
- Weight of the object (5,000 N) acts vertically downward.
- Tension force from Cable 1 (T1) acts horizontally to the right.
- Tension force from Cable 2 (T2) acts upward and to the left at an angle of 37.0° to the negative x-axis.
2. Analyze the vertical forces:
Since the object is in equilibrium (not accelerating vertically), the sum of the vertical forces must equal zero.
Vertical forces: T2 (upward) - Weight (downward) = 0
T2 - 5,000 N = 0
T2 = 5,000 N
3. Analyze the horizontal forces:
Since the object is in equilibrium (not accelerating horizontally), the sum of the horizontal forces must equal zero.
Horizontal forces: T1 (to the right) = 0
T1 = 0
Therefore, the tension T1 is 0 N.
To find the tension, T1 in cable 1, we need to analyze the forces acting on the weight in equilibrium.
First, let's list down the given information:
Weight, W = 5000 N
Angle between cable 2 and the negative x-axis, θ = 37.0°
Now, let's break down the forces acting on the weight:
1. Weight, W: It acts vertically downward with a magnitude of 5000 N.
2. Tension, T1: It acts horizontally to the right, opposing the weight. We need to find its magnitude.
3. Tension, T2: It acts upward and to the left at an angle of θ = 37.0° to the negative x-axis.
In equilibrium, the sum of all the forces in the x-direction and y-direction should be zero.
Let's start by analyzing the forces acting in the y-direction:
Sum of forces in the y-direction = T2 * sin(θ) - W = 0
T2 * sin(θ) = W
T2 * sin(37.0°) = 5000 N
T2 = 5000 N / sin(37.0°)
T2 ≈ 8164.26 N
Now, let's analyze the forces in the x-direction:
Sum of forces in the x-direction = T1 - T2 * cos(θ) = 0
T1 = T2 * cos(θ)
T1 = 8164.26 N * cos(37.0°)
T1 ≈ 6485.77 N
Therefore, the tension, T1 in cable 1 is approximately 6485.77 N.