Finding the Angle of Tipping for a Semi-Cylinder on a Ramp

In summary: I don't know what else to try...In summary, the student was unable to solve the problem due to not understanding the concept of phi. They were able to find the distance between the center of the cylinder and the point of contact between the ramp and cylinder using the equation r*(1-4/(3*pi)), but they were not able to find the torque due to the fact that the line of contact and the line of motion will be in different directions.
  • #1
PennyGirl
23
0
[/tex]

Homework Statement


http://emweb.unl.edu/negahban/em223/sexam3/sexam3.htm"

number 3...the one about the semicylinder on a ramp

Homework Equations


[tex]\Sigma[/tex] F_x = 0
[tex]\Sigma[/tex] F_y = 0
[tex]\Sigma[/tex] M=0
f = [tex]\mu[/tex] * N

The Attempt at a Solution


I got the first part to be 16.7 degrees. However, I can't find \phi I just need to be pointed in the right direction...
 
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  • #2
If you were to draw a line from G to the point of contact between the cylinder and ramp, phi occurs when that line coincides with the direction of mg.
 
  • #3
Gear300 said:
If you were to draw a line from G to the point of contact between the cylinder and ramp, phi occurs when that line coincides with the direction of mg.

So then I know that the distance between G and the ramp is r*(1-4/(3*pi)), but I don't know how that helps me solve the problem...
 
  • #4
PennyGirl said:
So then I know that the distance between G and the ramp is r*(1-4/(3*pi)), but I don't know how that helps me solve the problem...

I wouldn't say that would be the distance...that would be the case when the line from G to contact falls on a radial line, which wouldn't be the same case as when the line from G to contact falls on the vector mg.
 
  • #5
Gear300 said:
I wouldn't say that would be the distance...that would be the case when the line from G to contact falls on a radial line, which wouldn't be the same case as when the line from G to contact falls on the vector mg.

right...that makes sense. I'm still having a hard time picturing how this fact will help me solve the problem...
isn't the distance between G and the contact surface going to be in the same direction as mg, otherwise the semi-cylinder would try and move back to the place where they are in the same direction (because g will cause a torque?) sorry if that doesn't make sense...
 
  • #6
PennyGirl said:
...isn't the distance between G and the contact surface going to be in the same direction as mg, otherwise the semi-cylinder would try and move back to the place where they are in the same direction (because g will cause a torque?) sorry if that doesn't make sense...

Yup...that is what is going on...and the line covering that distance for this particular case does not necessarily coincide with a radial line (by radial line, I'm referring to the line from the center of what would have been the complete cylinder to the point of contact).
 
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  • #7
Gear300 said:
Yup...that is what is going on...and the line covering that distance for this particular case does not look like it would coincide with a radial line (by radial line, I'm referring to the line from the center of what would have been the complete cylinder to the point of contact).

So right before it is going to tip, torque = zero...maybe?
 

1. What is "Hemisphere on Ramp: Find Phi"?

"Hemisphere on Ramp: Find Phi" is a mathematical concept that involves finding the angle (represented by the Greek letter phi) at which a hemisphere is placed on a ramp in order to maintain equilibrium.

2. Why is "Hemisphere on Ramp: Find Phi" important?

This concept is important because it helps determine the optimal angle at which a hemisphere can be placed on a ramp without rolling or sliding, which is useful in engineering and architecture.

3. How is "Hemisphere on Ramp: Find Phi" calculated?

The calculation involves using trigonometric functions such as sine, cosine, and tangent to find the angle phi. The specific formula used depends on the angle of the ramp and the radius of the hemisphere.

4. What are the real-world applications of "Hemisphere on Ramp: Find Phi"?

This concept is used in various fields such as physics, engineering, and architecture to determine the optimal angle for placing a hemisphere on a ramp. It can also be applied to other situations involving equilibrium, such as balancing objects on an inclined surface.

5. Can "Hemisphere on Ramp: Find Phi" be applied to other shapes besides a hemisphere?

Yes, the concept can be applied to other shapes as long as they have a circular base and are placed on a ramp. The specific calculations may differ, but the overall concept remains the same.

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