Vector Calculus
posted by Lalo on .
A horizontal clothesline is tied between 2 poles, 20 meters apart.
When a mass of 5 kilograms is tied to the middle of the clothesline, it sags a distance of 4 meters.
What is the magnitude of the tension on the ends of the clothesline?

Requires algebra, trigonometry and physics, not calculus.
Total force down = total force up since the mass does not accelerate
Do half the problem at a time since it is symmetrical
So 2.5 kg down on left gives 2.5*9.8 = 24.5 Newtons down.
That 24.5 Newtons must be supported by the vertical component of the tension in the line on the left.
So T (4/sqrt (4^2+10^2) = 24.5
T = 24.5 (sqrt 116)/4 = 66 Newtons
If this is a mathematics rather than a physics question they may express force in kilograms (mathematicians do stuff like that) in which case divide by g which is about 9.8 m/s^2 
T 4/sqrt (4^2+10^2) DOES NOT EQUAL 24.5

ignore my last comment i'm an idiot