Statics Question II

[wildareal]

How do you prove that Triangle ACB is isoceles?

[Mao]

It is not necessarily isoceles.

Since the system is in equilibrium, the sum of forces must be zero. Also, from "using strings and pulleys", we can assume that point A and B are two frictionless pulleys.
Thus, define T1 to be the tension force in the wire 6kg-A-C, and T2 is the tension force in the wire C-B-8kg.
define the angle between AC and the horizontal to be , and the angle between BC and the horizontal to be (these angles are the angle between the blue line(s) and the solid black line)

Resolving forces:

6kg,
8kg,
10kg,
horizontal,
vertical,

Solving this on the 89 with domain restrictions (so the calc don't freak out)

Code: [Select]

solve(6cos(x)=8cos(y) and 6sin(x)+8sin(y)=10,{x,y})|0<x<pi/2 and 0<y<pi/2

The answer is



Thus,

[Mao]

For interest sake, I've done a little analysis on this (I was wondering whether ACB will always be 90 degrees).

If we denote each of the masses m1, m2 and m3 (from left to right), then eventually,





And if ACB is to be a right angle, then the three masses must follow , and the expression for the above simplifies to , funny how you see Pythagorean identity creep into every right angle triangles, eh?

[wildareal]

For interest sake, I've done a little analysis on this (I was wondering whether ACB will always be 90 degrees).

If we denote each of the masses m1, m2 and m3 (from left to right), then





And if ACB is to be a right angle, then the three masses must follow , funny how you see Pythagorean identity creep into every right angle triangles, eh?

Thanks Mao. Yer they do. I did it using a triangle (6,8,10) with angle b=90 degrees therefore ACB=190-90 degrees, but I'm not sure if its valid?