Moment about a drum (Answer Check)

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The discussion revolves around verifying the calculation of the moment about a drum using the equation M = Fd. The user calculated the moment with various forces and angles, arriving at an angle α of 39.1 degrees. Another participant suggested a different approach, emphasizing that relocating forces to the center of the drum should yield the same moment. They also calculated a different angle of 34.27 degrees, indicating a potential error in the original calculation. The conversation highlights the importance of checking calculations and considering alternative methods in physics problems.
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Homework Statement


See Figure.


Homework Equations


M = Fd


The Attempt at a Solution



I split all the forces into rectangular components and calculated their perpendicular distance from the point E.

M_{R}= 50 = (15.5 * 0.72) + (58 * 0.56) - (63.1 * 0.61) + (90.1 * 1.18) - (130sin \alpha * (0.75 + 0.75cos \alpha ) ) + (130cos \alpha * 0.75sin \alpha)

So,

\alpha = 39.1^{o}

Can someone verify my answer by chance?

Thanks again.
 

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Bump still looking for an answer check :biggrin:!

Anyone!?
 
The best check is to insert the result into the system described by equivalences picturing the same reality in a different way.
Notice that all the line of forces go through the center of the drum. If you relocate them there, the moment imposed on E won't change: the force is the same angle and magnitude, and the arm is also the same. Now let's compute the sum of the forces (The complex coordinate system is right to positive reals, up for positive imags, and I am just too addicted to express angles relative to due north.):
<br /> 120 - 60 \operatorname{sin}\left(-15\right) + 110 \operatorname{sin}\left(145\right) + \\<br /> 130 \operatorname{sin}\left(50.1\right) + 60 \mathbf{\imath} \operatorname{cos}\left(-15\right) + 110 \mathbf{\imath} \\<br /> \operatorname{cos}\left(145\right) + 130 \\<br /> \mathbf{\imath} \operatorname{cos}\left(50.1\right)
unfortunately imag(F1+F2+F3+F4)*0.75= 37.3775097537 which is not quite the 50 I would expect.

Either I am or you are in error:)
 
I found a much simpler way to solve, I found that alpha = 34.27 degrees.
 
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