1. 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?

2. 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.

$r_s = small\ radius$
$r_l = large\ radius$

3. 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

4. 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?

5. 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.

6. 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?

7. 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°.

8. Okay, I can see that. Thanks a lot for all your help!

9. Re: circle problem

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?

10. Re: circle problem

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

11. Re: circle problem

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:
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.

12. 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.

13. Re: circle problem

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:
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.

14. 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.
Hello,

in the 2nd drawing the two yellow angles must be equal. To prove that use the property that a radius is perpendicular to a tangent in the tangent point. In addition you can use similarities between right triangles formed by the tangents, the radii und the connection between the midpoints.

I've modified the 2nd drawing a little bit to show you what I mean:

The indicated triangles must be similar because both are right triangles (the radii are perpendicular to the same tangent) and for symmetry reasons the angle at the crosspoint of the tangents must be equal too. Therefore the 3rd angle in both triangles must be equal.

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how long must the belt be if the pulleys are rotating in opposite directions

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