# Effect of angle and lever length on moment

First, some background on my problem: I'm a sailor, and I'm looking to attach a rope between the base of my boat's mast and the boom. We refer to this rope as a boom vang, and its purpose is to pull the boom down. Because the boom may swing from side to side, the rope is attached at one end to the mast itself, in line with where the boom attached. The question is, where on the boom should the other end of the vang be attached to maximize the downward moment?

Here is a diagram (it's terrible - sorry), with the rope in solid blue:

The rope leads off to the sailor, who can apply tension to it, some of which will become a downward force. The other variable is the distance from the mast to where the vang attaches to the boom, L. Theta is the angle between the rope and the deck (or the boom).

It seems to me that the moment is the product of the downward force and the length of the lever that it's working on, L. Now, as L is increased, theta is decreased, reducing the downward force (compressing the boom instead). In the extreme cases, where theta is 90 degrees (parallel to the mast, thus maximum downward force, but no lever to act on), or infinitesimal (parallel to the infinitely long boom, providing a great long lever to act on but all of the force in the horizontal) there is no moment.

So, how to choose the correct angle? If I remember correctly (and I'm not sure that I do, which is why I'm here :)), the vertical component of the force should be sine(theta). Further, L should be cosine(theta). Since the moment is their product, the value for any theta should be sin(theta)*cos(theta). Is that correct? If so, the maximum should be theta = pi/4 radians, correct?

Quick disclaimer: I haven't practiced this sort of math in nigh 20 years.

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You are right that you want to maximize is the torque around the point where the boom attaches to the mast. If $F$ is the force in the vang, that torque is the projection of that force on a vector perpendicular to the boom, which is downward. The downward force is then $F \sin \theta=F \frac H{H^2+L^2}$ and the moment is then $LF=F\frac {LH}{H^2+L^2}$, which for a given $H$ is maximized when $H=L$
@fatcat1111: That is a restating of the same solution. Because the boom is parallel to the deck, the angle between the boom and vang is $\theta$. Then $\tan \theta = \frac HL$ so the $L$ I used in the moment is the same as $\frac H{\tan \theta}$ –  Ross Millikan Nov 12 '12 at 20:31
Your calculation is correct, as far as the force calculation goes. It's useful to note that $\sin(\theta)\cdot\cos(\theta)=\frac12\sin(2\theta)$, and that is maximal when $\theta=\pi/4=45^\circ$. Thus $L=H$ is optimal.