A crane suspends a 500lb steel beam horizontally by support cables (of negligible weight) attached from a hook to each end of the beam. The support cables make an angle of 60 degrees with the beam. Find the tension vector in each support cable and the magnitude of each tension.

The sum of forces in a system in equilibrium is zero

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

∑ = 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 = ≈ 288.7

The magnitude of the tension in each support cable, T₁ =  ≈ 288.7 lb

The tension in vector form:

The tension in each support cable in vector is expressed as follows

= (500/√3) × sin(60°)·i +  (500/√3) × cos(60°)·j

Which gives;

= (500/√3) × (√3)/2·i +  (500/√3) × 1/2·j = 250·i + 250·(√3)/3·j

Similarly, we have;

= (500/√3) × sin(60°)·i - (500/√3) × cos(60°)·j

Therefore:

Learn more about equilibrium of forces here:

link

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 distance from any point on the parabola to the focus is the exact same as the distance from that point to the directrix.

if you could provide answer choices that might allow me to answer the question easier if my above answer isn't an option! (:

equal to

step-by-step explanation: