1. 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)

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

2. 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)

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 $\displaystyle 147.48^{\circ}$ for angle AOB and $\displaystyle 212.52^{\circ}$ 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.

3. Hello, Educated!

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

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

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

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

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

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

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

Can you finish up?

4. 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.

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

$\displaystyle Length = 99.1cm$