Respuesta :
Answer:
T1 = sqrt(250^2 + (250/sqrt(3))^2)
T2 = sqrt(250^2 + (-250/sqrt(3))^2)
The vectors are the two expressions int he square root, basically sqrt(y^2 + x^2)
Step-by-step explanation:
Maybe should be n the physics category
Make a free body diagram that is the triangle witht he two 60 degree angles, so the third is also 30 degrees, the tension forces pointing upward along the two sides and a weight vector pointing down.
You have to break each tension force into its vertical and horizontal components, essentially split that equilateral triangle in half. The to share the vertical force but have equal and opposite horizontal forces.
I will call the weight W, and the tensions T1 and T2. Also, splitting them into their horizontal and vertical components I will call those T1x, T1y, T2x and T2y.
Essentially you have two right triangles you know one leg of each, and an angle. the leg you know is T1y and T2y. We also know these two add up to 500, and the othr two legs add up to 0. so T1x + T2x = 0 or T1x = T2x.
Trig time. Lets take triangle 1 (with T1, T1x and T1y) using tangent we get tan(60) = T1y / T1x so to solve for T1x we get T1x = T1y / tan(60) = T1y/sqrt(3). Similarly for T2x we get T2y/sqrt(3). Since T2x is in the opposite direction of T1x though this is negative so T2x = -T2y/sqrt(3)
Now using T1x = -T2x we get T1y/sqrt(3) = T2y/sqrt(3) or T1y = T2y
Then using T1y + T2y = 500, or specifically T1y = 500- T2y we get
(500- T2y) = T2y
500 = 2*T2y
T2y = 250
Then backsubstituting we get
T1y = 250
T2x = -250/sqrt(3)
T1x = 250/sqrt(3)
Then use the pythagorean theorem to find T1 and T2
T1 = sqrt(250^2 + (250/sqrt(3))^2)
T2 = sqrt(250^2 + (250/sqrt(3))^2)
The sum of forces in a system in equilibrium is zero
- The tension vector acting in each support cable are;
[tex]\mathbf{\overset \longrightarrow {T_1}} = \left(250 \cdot \mathbf{i} + \dfrac{250 \cdot \sqrt{3} }{3} \cdot \mathbf{j}\right) lb[/tex]
[tex]\mathbf{\overset \longrightarrow {T_2}} = \left(250 \cdot \mathbf{i} - \dfrac{250 \cdot \sqrt{3} }{3} \cdot \mathbf{j}\right) lb[/tex]
- The magnitude of each tension is [tex]\dfrac{500 \cdot \sqrt{3} }{3} \ lb[/tex]
The reason the above values are correct is as follows:
Question: Please find attached a diagram of the beam obtained from a similar question
The given parameters of the crane are;
The weight of the beam, W = 500 lb
The orientation of the beam = Horizontal
The angles the supporting cables make with the beam, = 60°
Required:
To find the vector form of the tension in each supporting cable
Solution:
Let T₁ and T₂ represent the tension in each supporting cable, we have;
At equilibrium, the sum of forces is zero
∑[tex]\mathbf{F_y}[/tex] = 0
Therefore;
W - (T₁ × sin(θ) + T₂ × sin(θ)) = 0
W = (T₁ × sin(θ) + T₂ × sin(θ))
Therefore;
500 = (T₁ × sin(60°) + T₂ × sin(60°))
We also have, ∑Fₓ = 0
Therefore;
(T₁ × cos(θ) - T₂ × cos(θ)) = 0
T₁ × cos(θ) = T₂ × cos(θ)
Therefore;
T₁ × cos(60°) = T₂ × cos(60°)
∴ T₁ = T₂, which gives;
500 = (T₁ × sin(60°) + T₂ × sin(60°)) = (T₁ × sin(60°) + T₁ × sin(60°))
500 = (T₁ × sin(60°) + T₁ × sin(60°)) = 2 × T₁ × sin(60°)
T₁ = 500/(2 × sin(60°)) = 500/√3 = [tex]\dfrac{500 \cdot \sqrt{3} }{3}[/tex] ≈ 288.7
The magnitude of the tension in each support cable, T₁ = [tex]\dfrac{500 \cdot \sqrt{3} }{3} \ lb[/tex]≈ 288.7 lb
The tension in vector form:
The tension in each support cable in vector is expressed as follows
[tex]\overset \longrightarrow {T_1}[/tex] = (500/√3) × sin(60°)·i + (500/√3) × cos(60°)·j
Which gives;
[tex]\overset \longrightarrow {T_1}[/tex] = (500/√3) × (√3)/2·i + (500/√3) × 1/2·j = 250·i + 250·(√3)/3·j
[tex]\mathbf{\overset \longrightarrow {T_1}} = \left(250 \cdot \mathbf{i} + \dfrac{250 \cdot \sqrt{3} }{3} \cdot \mathbf{j}\right) lb[/tex]
Similarly, we have;
[tex]\overset \longrightarrow {T_2}[/tex] = (500/√3) × sin(60°)·i - (500/√3) × cos(60°)·j
Therefore:
[tex]\mathbf{\overset \longrightarrow {T_2}} = \left(250 \cdot \mathbf{i} - \dfrac{250 \cdot \sqrt{3} }{3} \cdot \mathbf{j}\right) lb[/tex]
Learn more about equilibrium of forces here:
https://brainly.com/question/21483623
