Cable Tension

Section 6.3

Please, do 83 in step by step fashion.

 

83.

let cable between left crane and weight be t1
and cable between right crane and weight be t2
drawing right triangles on both sides of weight, we see that horizontal distance fron the weight left is
t1*cos(24.3) =x =>cos(24.3) =x/t1
and horizontal distance (x value) fron the weight righ is
t2*cos(44.5)=x =>cos(44.5)=x/t2

note: is not equal to

vertical distance (y value) on the left side is
t1*sin(24.3)=y=>sin(24.3)=y/t1

vertical distance (y value) on the right side is
t2*sin(44.5) =y => sin(44.5) =y/t2
note: vertical distances is not equal to each other


but horizontal distances are same, so

t1*cos(24.3) =t2*cos(44.5)

t1*cos(24.3) -t2*cos(44.5)=0

given the sum of these two forces

t1*sin(24.3) +t2*sin(44.5)=20240

so we have system of linear equation to solve:

t1*cos(24.3) -t2*cos(44.5)=0

t1*sin(24.3) +t2*sin(44.5)=20240
--------------------------------------------------

solving it we get

t1≈15484.094lb , t2≈19785.832b

 

I sure wish I had your math skills. It must be wonderful to see a problem that you have not seen and know exactly what to do. Knowing the WHY in terms of lecture notes is super important.