Center of Mass involving Dumbbell With Uneven Weigths

[DebonDevil]

Homework Statement

You work in a plant that manufactures heavy
dumbbells. Due to a manufacturing error, one side of a
100 kg dumbbell was found to be 60 kg (M1) while the
other was only 40 kg (M2). The mass of the bar itself is
negligible. The factory has provided you and your
colleague with special handles that permit one to lift one
side of the dumbbell directly above the center of mass of
that side. You and a colleague are asked to hand carry the
dumbbell to the analysis lab on the other side of the
compound. You have been assigned the heavy side (M1).
You believe that if you lift your end of the dumbbell higher
than your colleague, it will result in a fair, even distribution
of weight. At what height above the center of mass of the
opposite end of the dumbbell must you lift the center of
mass of your end your end for this to be so? At what angle
with respect to x does this occur?

Homework Equations

C_mass = $\frac{M1*(L1)+ M2*(L1+L2)+M3*(L1+L2+L3)}{M1+M2+M3}$
$\tau$ = F*d

The Attempt at a Solution

Setting up an equation where the two torques around the center of mass (which I got as .4 from M1 and .6 from M2) are equal to 0 in order to find the $\theta$ that would result in find the "y" value for how high we'd have to lift M1 in order to make it even.:

0 = 60(.4)cos(180-$\theta)$ - 40(.6)cos($\theta$)
0= 60(.4)sin(180-$\theta)$ - 40(.6)

My friend assisted me with this problem, however, I don't fully understand why he arbitrarily chose to use cosine (I know that a right-triangle is involved, just not HOW) or why the first cosine is 180°-$\theta$.

 

[member 553083]

I also don't understand why the second equation is the same as the first, only with sine instead of cosine.The reason your friend used cosine and sine is because they are both trigonometric functions that help to define angles and side lengths in a right triangle. In this case, you have two objects (the heavy dumbbell and the light dumbbell) separated by a distance. The angle between them (θ) can be determined by using the cosine and sine functions.The cosine formula is used to determine the angle between two sides of a right triangle, given their lengths. For this problem, the length of one side is the distance between the center of mass of the two dumbbells, and the length of the other side is the distance between your end of the dumbbell and the center of mass of the light dumbbell.The sine formula is similar to the cosine formula, but it is used to determine the length of one side of a triangle, given the angle and the length of the other side. For this problem, the length of the side you are trying to find is the height at which you must raise your end of the dumbbell in order for the two torques around the center of mass to be equal and opposite.The reason the second equation is the same as the first, but with sine instead of cosine, is because the two equations are related since they both involve the same angle (θ). So, if you know one of the equations, you can use it to solve for the other.