# Applied pure trig question

• Feb 8th 2010, 04:11 AM
dojo
Applied pure trig question
A belt ABCDEFA of length L passes round two wheels, centres O_1 and O_2. The parts ABC and DEF of the belts are in contact with the wheels and parts AF and CD are straight. If O_1A = a and O_2F = 3a and the angle AO_1B = theta radians show that:

4atan (theta) = 4a(theta) + L - 6a*pi

Next, find the shortest length belt in terms of a.

Im pretty stuck on this one. Any ideas??
I am looking at the equation to see what its saying but can't see where tan(theta) comes from - Obviously a triangle is drawn and the Opposite and Adjacent sides are used some how.
The arc AC on the small wheel is 2a(2*pi - (theta)) but I'm just kinda stuck
• Feb 8th 2010, 05:20 AM
Hello dojo
Quote:

Originally Posted by dojo
A belt ABCDEFA of length L passes round two wheels, centres O_1 and O_2. The parts ABC and DEF of the belts are in contact with the wheels and parts AF and CD are straight. If O_1A = a and O_2F = 3a and the angle AO_1B = theta radians show that:

4atan (theta) = 4a(theta) + L - 6a*pi

Next, find the shortest length belt in terms of a.

Im pretty stuck on this one. Any ideas??
I am looking at the equation to see what its saying but can't see where tan(theta) comes from - Obviously a triangle is drawn and the Opposite and Adjacent sides are used some how.
The arc AC on the small wheel is 2a(2*pi - (theta)) but I'm just kinda stuck

What you don't say - but what I deduce from your comment about the arc $\displaystyle AC$ - is that $\displaystyle BO_1O_2E$ is a straight line.

So look at the attached diagram. The key fact that you seem to be missing is that when we draw $\displaystyle O_1G$ parallel to $\displaystyle AF,\; O_1O_2G$ is a right-angled triangle with $\displaystyle \angle O_1O_2G = \theta$ and $\displaystyle O_2G = 2a$. So $\displaystyle O_1G = AF = 2a\tan\theta = CD$.

So now add the arc lengths $\displaystyle ABC$ and $\displaystyle D EF$ to $\displaystyle AF + CD\; (=4a\tan\theta)$, equate the sum to $\displaystyle L$, and the result follows.

The shortest belt possible is when $\displaystyle O_1O_2 = a + 3a = 4a$. This gives, from $\displaystyle \triangle O_1O_2G, \;\cos\theta = \tfrac12$. So with this value of $\displaystyle \theta$, solve the equation for $\displaystyle L$ and you're there.

Can you complete it now?