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- #1

I set the problem like this:

And I thought the side labeled 'x' is the force needed to keep the wheel up there, which, if calculated by sin(3) = 600/x is 1,200 lbs. but that's not the correct solution.

What am I doing wrong here?

- Thread starter daigo
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- Thread starter
- #1

I set the problem like this:

And I thought the side labeled 'x' is the force needed to keep the wheel up there, which, if calculated by sin(3) = 600/x is 1,200 lbs. but that's not the correct solution.

What am I doing wrong here?

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- Jan 26, 2012

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Gravity is pulling the object straight downwards, and the ramp is affecting the gravity's pull which makes the weight roll down the ramp, so the two forces are gravity pulling the object down, and the object itself going down the ramp. But this is essentially the same as how I depicted it initially, isn't it?

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- Jan 26, 2012

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- Jan 26, 2012

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Apologies to Ackbach if this is giving away more than he intended but this page I found really nicely shows the whole scenario and how that leads to the various formulas.

http://zonalandeducation.com/mstm/physics/mechanics/forces/inclinedPlane/inclinedPlane.html

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- Jan 26, 2012

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I'm glad for you that you don't have to do any problems with inclines and friction. These problems can get really tedious, or at least they were for me when taking physics in high school and college.

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- Jan 26, 2012

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