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

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The discussion focuses on calculating the angle of tipping for a semi-cylinder on a ramp, with initial findings suggesting an angle of 16.7 degrees. Participants explore the relationship between the center of gravity (G), the point of contact on the ramp, and the gravitational force (mg). There is confusion regarding the distance from G to the ramp and its relevance to solving the problem, particularly in relation to torque. It is clarified that the line from G to the contact point does not align with a radial line when considering tipping. The consensus is that at the tipping point, the torque is effectively zero.
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Homework Statement


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

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

Homework Equations


\Sigma F_x = 0
\Sigma F_y = 0
\Sigma M=0
f = \mu * 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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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.
 
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...
 
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.
 
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...
 
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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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?
 
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