# Cog and Belt problem

• Sep 2nd 2010, 02:47 AM
Educated
Cog and Belt problem
There are 2 cogs, connected together by 1 belt. Find the length of the belt.

The large cog has a radius of 11cm. The small cog has a radius of 4cm. The distance between the centre of the 2 cogs is 25cm.
(Diagram not drawn to scale)

http://www.mathhelpforum.com/math-he...og-problem.png

I'm stuck because I cannot solve the angle AOB or YPZ, and I cannot solve the length of AY or BZ. Once I get those, then I can get solve the answer.

Can anyone tell me how to get the angle or length?
Does the angle AOB = angle YPZ?

I got $147.48^{\circ}$ for angle AOB and $212.52^{\circ}$ for the reflex angle of YPZ because I assumed they were equal, but it doesn't seem right.
• Sep 2nd 2010, 04:29 AM
mr fantastic
Quote:

Originally Posted by Educated
There are 2 cogs, connected together by 1 belt. Find the length of the belt.

The large cog has a radius of 11cm. The small cog has a radius of 4cm. The distance between the centre of the 2 cogs is 25cm.
(Diagram not drawn to scale)

http://www.mathhelpforum.com/math-he...og-problem.png

I'm stuck because I cannot solve the angle AOB or YPZ, and I cannot solve the length of AY or BZ. Once I get those, then I can get solve the answer.

Can anyone tell me how to get the angle or length?
Does the angle AOB = angle YPZ?

I got $147.48^{\circ}$ for angle AOB and $212.52^{\circ}$ for the reflex angle of YPZ because I assumed they were equal, but it doesn't seem right.

Draw a line parallel to BZ from P to the radius OB. Let it meet OB at Q. Note that QO = 7. Use the right-triangle QOP to get:

1. The angle QPO and hence the angle OPZ.
2. The angle QOP.
• Sep 2nd 2010, 08:01 AM
Soroban
Hello, Educated!

Your diagram didn't show up, but I'll take a guess.

Quote:

There are 2 cogs, connected together by a belt.
Find the length of the belt.

The large cog has a radius of 11cm. The small cog has a radius of 4cm.
The distance between the centres of the 2 cogs is 25cm.

I got . $\angle AOB = 147.48^o$ and . $\angle YPZ = 212.52^o$
These happen to be correct.

In the diagram, consider quadrilateral $OPZB.$
. . Note that it has two right angles.

Code:

    O *       | @ *     7 |      *  25       |          *       |              *     Q * - - - - - - - - - * P       |        24        |     4 |                  | 4       |                  |       |                  |     B * - - - - - - - - - * Z               24

In right triangle $OQP\!:\;QP^2 + 7^2 \:=\:25^2 \quad\Rightarrow\quad QP = BZ = 24$

Also: . $\sin\theta \:=\:\frac{24}{25} \quad\Rightarrow\quad \theta \:\approx\:73.74^o$

and: . $\angle OPZ \:=\:180^o - \theta \:=\:106.26^o$

Can you finish up?
• Sep 2nd 2010, 10:28 PM
Educated
Oh, I got it.

The only reason I wasn't getting the right answer was because I forgot to use the pythagorus theorum on the straight length of the belt.

$Length = 22\pi * \frac{212.52}{360} + 8\pi * \frac{147.48}{360} + 24 * 2$

$Length = 99.1cm$