1. Feb 3, 2010

### jegues

1. The problem statement, all variables and given/known data
See Figure.

2. Relevant equations
$$M = Fd$$

3. 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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2. Feb 4, 2010

### jegues

Bump still looking for an answer check !

Anyone!?

3. Feb 6, 2010

### magwas

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.):
$$120 - 60 \operatorname{sin}\left(-15\right) + 110 \operatorname{sin}\left(145\right) + \\ 130 \operatorname{sin}\left(50.1\right) + 60 \mathbf{\imath} \operatorname{cos}\left(-15\right) + 110 \mathbf{\imath} \\ \operatorname{cos}\left(145\right) + 130 \\ \mathbf{\imath} \operatorname{cos}\left(50.1\right)$$
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:)

4. Feb 7, 2010

### jegues

I found a much simpler way to solve, I found that alpha = 34.27 degrees.

Last edited: Feb 8, 2010