(torque) a force needed for wheel to rise up onto box

[slambert56]

ok this was a problem in my book and the answer apparently is 29N.

I have the diagram and pictures below. I know that the pivot point will be the edge of the block that is touching the outside of the wheel. When the wheel is in equilibrium the sum of the torque will=zero so i can set up the equation where the sum t=force times lever arm(l). When the wheel rises the normal force will equal zero. I am having major trouble getting the lever arms though. I know how to do vectors but I don't have any angles to work with so I am just trying the use the radius. Any guidance would be greatly appreciated.
http://img7.imageshack.us/img7/3214/mmspicture1p.jpg

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here is my diagram
http://img641.imageshack.us/img641/7379/001aoz.jpg

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[Danger]

I don't know anything about math, but it seems to me that the question can't be answered because critical information is missing. I refer specifically to the frictional connection between the wheel and the obstacle. You could put a couple of thousand Newtons into that wheel, but if there's no grip it will just sit there and spin.

 

[slambert56]

I think it is one of those frictionless problems. I forgot to post the actual question ha.
the drawing shows a bicycle wheel resting against a small step (height=.120m). The weight and radius of the wheel are w=25N and r=.340m. A horizontal force F is applied to the axle of the wheel. As the magnitude of F increases there comes a time when the wheel just begins to rise up and lose contact with the ground. What is the magnitude of the force when this happens.

 

[Danger]

Sorry, man, but I'm lost. First off, I have a grade 9 math level. Secondly, there were never such things as frictionless questions when I was studying science in school. Everything was "real world". Ideal machines didn't exist. We measured stuff, with all imperfections included.

 

[Red Kangaroo]

Use unit circle formula to determine the angle between the applied force and the pivot point.

Then do your summation of torques to zero. Two forces acting. Weight & Pull.

T = r X F cross product

So

T = rF sin θ

Answer comes to 29N.

 

[Doc Al]

I have the diagram and pictures below. I know that the pivot point will be the edge of the block that is touching the outside of the wheel. When the wheel is in equilibrium the sum of the torque will=zero so i can set up the equation where the sum t=force times lever arm(l). When the wheel rises the normal force will equal zero. I am having major trouble getting the lever arms though. I know how to do vectors but I don't have any angles to work with so I am just trying the use the radius.

You should be able to get the moment arm of the applied force F without any angles--just subtract.

As for the moment arm of the weight, consider a right triangle where the hypotenuse is the line from the wheel axis to the step edge. You should know two of the three sides (use the result from the above calculation). A little trig will get you the third.

 

[slambert56]

Thanks red and Doc I was making things more complicated then they should have been. It seemed weird at first that the normal force and push force had the same lever arm but I see it now.

http://img847.imageshack.us/img847/4014/001hxy.jpg

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finally got it.

 

[Red Kangaroo]

You should be able to get the moment arm of the applied force F without any angles--just subtract.

As for the moment arm of the weight, consider a right triangle where the hypotenuse is the line from the wheel axis to the step edge. You should know two of the three sides (use the result from the above calculation). A little trig will get you the third.

Yes you can do it without angles, but I did it without hypotenuse...

More than one way to skin a cat!