Thread: 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

Hello dojo

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 - is that is a straight line.

So look at the attached diagram. The key fact that you seem to be missing is that when we draw parallel to is a right-angled triangle with and . So .

So now add the arc lengths and to , equate the sum to , and the result follows.

The shortest belt possible is when . This gives, from . So with this value of , solve the equation for and you're there.

Can you complete it now?

Grandad

Perfect. It's kind of abstract but I see it so clearly now. Many thanks