Tension in each cable physics 101?

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Hello

This can not be solved, unless you have an angle between the cables, or of the cables with vertical or horizontal. or distance between the cables. The cables on your picture can have any positions.
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Ok, take the angle between the red cable and the vertical, and the horizontal given as 45°.
Then is the distance from the point where the 3 cables meet, to the ceiling = x = 9*sin45, and
the angle of the other (black) cable with the vertical is cos^-1 ( x/11) = 9sin45/11 = 54.65°.
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To find the forces in the cables draw a parallelogram of forces:
elongate the yellow cable upwards, and mark a distance from the uniting point of the cables of 1000 lbs. ( = 10 cm up, for instance , --> 1 cm = 100 lbs). Now draw parallels to the cables through this point. --> you get a parallelogram. The sides of the parallelogram are the forces in the cables with 1 cm = 100 lbs.

You can calculate the magnitudes of the forces (i.e., the lengths of the sides of the parallelogram ) using the sine rule for triangles:
the 2 congruent triangles (the parallelogram is divided by the vertical force vector into two congruent triangles) have angles 45°, 54.65°, and 80.34°.

and 1000/sin80.34 = (Force red)/(sin54.65) = (Force black)/(sin45)
Force red = 1000*sin54.65/sin80.34 = 827.36 lb
Force black = 1000*sin45/sin80.34 =712.27 lb

the force in the yellow cable is always = 1000 (lb)
(is lb a force or a mass? I am not familiar with the lb/ft system, anyway, you know what to do)

I hope it helps. I you do not understand me, mail me

Regards...Show more

maximum suitable so some distance Sum forces in the horizontal on the element the place all 3 cables meet TR*cos30-TL*cos60=0 And in the vertical TR*sin30+TL*sin60-197.8=0 resolve the equations for the TR and TL...Show more