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Section 6.3
Please, do 83 in step by step fashion.
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
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solving it we get
t1≈15484.094lb , t2≈19785.832b