Why is this the torque

I know I'm missing something simple..

a man is holding an 8.00 kg vacuum cleaner at arm's length, a distance of 0.55m from his shoulder. what is the torque on the shoulder joint if the arm is held at an angle of 30 degrees below the horiz.?
answer is 37.3 N.m

but why?? I got 21.6 :/

8.00(kg)*9.81(m/(s*s)) = 78.48 N.
Torque = FXr. ==> 78.48 N * 0.55m = 43.164 N*m. (if held horizontal).

Where you went astray was when you used sin(30) * 43.164 N*m = 21.6 Nm
You need to use cos(30), then you will get 37.3 N*m.

The reason you use cos(30) instead of sin(30) is because it is 30 deg from horizontal, this means that your "lever arm" is cos(30)*radius away from the shoulder. (or the x dimension)

In vectors,
τ_ = r_ x F_ = r_ x (-mgz_)
And the r and (-z) vectors are separated by an angle of 60º. So that makes it
τ = ||τ_|| = mgr sin60º = mgr cos30º

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Looks like you multiplied by the sine of 30° when you should've done the cosine of 30°. The reason why it is the cosine of 30° is because you want the vector of the vacuum's weight that is perpendicular to the arm, not parallel.

The weight of the vacuum acts on a vertical line. The horizontal distance of this line from the shoulder is H = L*cos30° = .55*cos30° = .4763 m
So,
Q = W*H = m*g*H = 8.00*9.8*.4763 = 37.34 N∙m

That's why