Q.346. Static equilibrium.

Question 346.
Determine the distance y to the pulley for equilibrium. Neglect the size of the pulley. The pail and it's contents have a mass of 60 kg. If the cable BAC is 15 m long, determine the distance y to the pulley for equilibrium. Neglect the size of the pulley.

Answer 346.
Let T = tension in the cable

and AC = x
=> AB = 15 - x
Let AC make an angle α with the horizontal
and AB make an angle β with the horizontal

For equilibrium, balancing horizontal components of forces,
Tcosα = Tcosβ
=> √(x^2 - y^2) / x = [10 - √(x^2 - y^2)] / (15 - x)
=> (15 - x) √(x^2 - y^2) = 10x - x √(x^2 - y^2)
=> 15 √(x^2 - y^2) = 10x
=> 225 (x^2 - y^2) = 100x^2
=> x^2 = (9/5) y^2 ... ( 1 )

By geometry of the figure,
(y - 2)^2 + [10 - √(x^2 - y^2)]^2 = (15 - x)^2
=> y^2 - 4y + 4 + 100 - 20√(x^2 - y^2) + x^2 - y^2 = 225 - 30x + x^2
=> 104 - 4y - 20√(x^2 - y^2) = 225 - 30x
=> (30x - 4y - 121)^2 = 400 (x^2 - y^2)

Plugging x^2 = (9/5)y^2,
=> [30 * y √(9/5) - 4y - 121]^2 = 400 [(9/5) y^2 - y^2]
=> (36.25y - 121)^2 = 320 y^2
=> 36.25 y - 121 = ± 17.89y ... [negative root gives unacceptable result]
=> y = 2.235 m or 6.59 m
Acceptable value
y = 6.59 m.

[Note that the two right triangles with AB and AC as hypotenuse have to be similar as angle α = β.
Also, note that the answer is independent of the mass of the pail and its contents.]

Link to YA!