Question

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.

Answer 1

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)