# circle problem

• Sep 29th 2008, 08:26 PM
yoleven
circle problem
A narrow belt is used to drive a 50 cm diameter pulley from a 10 cm diameter pulley. The centers of the pulleys are 50 cm apart.
a.How long must the belt be if the pulleys are rotating in the same direction?
b How long must the belt be if the pulleys are rotating in the opposite direction...(the belt crosses over itself between the two pulleys)?
a. one pulley is 50 cm so its radius is 25cm
the other pulley is 10 cm so its radius is 5 cm
they are 50 cm apart from their centers.
50+(25+5)=80
the belt needs to be 80 cm long. Is that correct?
I'm not sure how to do part b of this question.

In addition, does anyone know where I can read more about situations like this so I can understand the situation better?
• Sep 30th 2008, 12:53 AM
earboth
Quote:

Originally Posted by yoleven
A narrow belt is used to drive a 50 cm diameter pulley from a 10 cm diameter pulley. The centers of the pulleys are 50 cm apart.
a.How long must the belt be if the pulleys are rotating in the same direction?
b How long must the belt be if the pulleys are rotating in the opposite direction...(the belt crosses over itself between the two pulleys)?
a. one pulley is 50 cm so its radius is 25cm
the other pulley is 10 cm so its radius is 5 cm
they are 50 cm apart from their centers.
50+(25+5)=80
the belt needs to be 80 cm long. Is that correct? No
I'm not sure how to do part b of this question.

In addition, does anyone know where I can read more about situations like this so I can understand the situation better?

Unfortunately I haven't much time yet so I only can give you the sketches to questions a) and b).

The belt is drawn in red. The triangles marked in red are right triangles whose dimensions are given so you are able to calculate distances (by Pythagorean theorem) and angles (by Cosine) which you need to calculate the arcs on the pulleys.

$r_s = small\ radius$
$r_l = large\ radius$
• Sep 30th 2008, 01:51 AM
earboth
Quote:

Originally Posted by yoleven
A narrow belt is used to drive a 50 cm diameter pulley from a 10 cm diameter pulley. The centers of the pulleys are 50 cm apart.
a.How long must the belt be if the pulleys are rotating in the same direction?
...

Here I am again:

I'm referring to my sketch #1 in my previous post.

1. The length of the sides of the right triangle are:

80 cm: distance between the midpoints of the pulleys
40 cm : difference of the 2 radii
therefore $s = \sqrt{80^2-40^2} = 40\cdot\sqrt{3}$

2. The interior angle at the midpoint of the large pulley is $\phi$:

$\cos(\phi) = \frac{40}{80}=\frac12~\implies~\phi=60^\circ$

Therefore the central angle of the arc $a_l$ is $\alpha= 240^\circ$ (Why?)
and the central angle of the arc $a_s$ is $\beta = 120^\circ$

3. The length of the belt is calculated by:

$L_{belt} = \underbrace{2\cdot 40\cdot\sqrt{3}}_{straight\ lines}+\underbrace{\frac{240}{360}\cdot 2\pi\cdot 50}_{large\ arc} + \underbrace{\frac{120}{360}\cdot 2\pi\cdot 10}_{small\ arc} = 80\cdot\sqrt{3} + \frac{220}3 \pi \approx 368.95\ cm$

to #b) Use my sketch to do this problem:
1. Calculate the straight line s
2. Calculate the angle $\alpha$
3. Calculate the length of the arcs

I've got as total length of the belt: 396.09 cm
• Sep 30th 2008, 05:56 AM
yoleven
Wow. Thanks. Those are really good sketches.
I have a few questions but the first one is, when you are figuring out the side of the right triangle in sketch #1, How do you know to take 40cm, (the difference of the two radi)?
How do you know that"s what you are going to do?
• Sep 30th 2008, 06:16 AM
earboth
Quote:

Originally Posted by yoleven
Wow. Thanks. Those are really good sketches.
I have a few questions but the first one is, when you are figuring out the side of the right triangle in sketch #1, How do you know to take 40cm, (the difference of the two radi)?
How do you know that"s what you are going to do?

1. The straight parts of the belt must be tangent to both circles. The tangent in question is translated so that it passes through the midpoint of the smaller circle and the larger circle is "shrunken"(?) by the length of the radius of the smaller circle.
If you look at those circles drawn by a dotted line then you have the this situation.

2. A tangent to a circle is perpendicular to the radius at the tangent point. I used this property to construct the right triangles which you need to calculate all missing values.
• Sep 30th 2008, 10:53 AM
yoleven
Okay, that makes sense to me. I can see how you got an angle of 240° in the larger wheel but I cant see how you got 120° in the smaller wheel. Could you explain this?
• Sep 30th 2008, 11:02 AM
earboth
Quote:

Originally Posted by yoleven
Okay, that makes sense to me. I can see how you got an angle of 240° in the larger wheel but I cant see how you got 120° in the smaller wheel. Could you explain this?

The radii of the 2 circles must be parallel because they are perpendicular to the same tangent. That means you have 2 parallels which are crossed by a straight line which connects the 2 midpoints. Now examine the angles at the midpoints: alternate angles, corresponding angles, etc.

As a result you'll get that the 2 yellow marked angles add up to 360°. So if one angle is 240° the second one must be 120°.
• Sep 30th 2008, 12:08 PM
yoleven
Okay, I can see that. Thanks a lot for all your help!
• Sep 28th 2014, 08:04 PM
andy1219
Re: circle problem
Quote:

Originally Posted by earboth
Here I am again:

I'm referring to my sketch #1 in my previous post.

1. The length of the sides of the right triangle are:

80 cm: distance between the midpoints of the pulleys
40 cm : difference of the 2 radii
therefore $s = \sqrt{80^2-40^2} = 40\cdot\sqrt{3}$

it said that The centers of the pulleys are 50 cm apart. How did you get the distance between the midpoints of the pulleys as 80cm?
• Sep 30th 2015, 12:01 PM
Re: circle problem
In b) i get how you found s and angle \alpha but how do you find \beta? Are both angles equal?
• Sep 30th 2015, 01:00 PM
earboth
Re: circle problem
Quote:

In b) i get how you found s and angle \alpha but how do you find \beta? Are both angles equal?

Good evening,

that's what I wrote:
Quote:

The radii of the 2 circles must be parallel because they are perpendicular to the same tangent. That means you have 2 parallels which are crossed by a straight line which connects the 2 midpoints. Now examine the angles at the midpoints: alternate angles, corresponding angles, etc.

As a result you'll get that the 2 yellow marked angles add up to 360°. So if one angle is 240° the second one must be 120°.
If this answer doesn't satisfy you please tell me where you're stuck.
• Sep 30th 2015, 01:06 PM
Re: circle problem
I understand how the 240 and 120 work for part a. But in part b, your yellow angles are similar in size. In part B, I found the theta angle and did 360-(2*theta) to get the alpha angle. Would I do the same as in part a next? I.e 360-alpha to get beta? Because i wasn't sure as I mentioned, the yellow angles that you have drawn look similar in size. Hopefully that explains my situation.
• Sep 30th 2015, 01:11 PM
earboth
Re: circle problem
Quote:

Originally Posted by andy1219
it said that The centers of the pulleys are 50 cm apart. How did you get the distance between the midpoints of the pulleys as 80cm?

Good evening,

the original wording of the question is:
Quote:

one pulley is 50 cm so its radius is 25cm
the other pulley is 10 cm so its radius is 5 cm
they are 50 cm apart from their centers.
50+(25+5)=80
The pulleys are 50 cm apart but you have to add the two different radii to get the distance between the midpoints.
• Sep 30th 2015, 01:18 PM
earboth
Re: circle problem
Quote: