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Mar 7 2019 10:27pm
Main Issue I am having is figuring out the horizontal Force required to stop the box. Tbh, I didn't think a Horizontal Force could stop the Vertical part of a force and therefore assumed the Horizontal force would just be COS() * Fnet. But that answer was wrong as well. (5.58)
Problem


What I have:



This post was edited by kasey21 on Mar 7 2019 10:31pm
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Mar 9 2019 06:06pm
find the force of the box moving down the ramp. This will include the component of its gravity down the ramp and the frictional force up the ramp.
(Fg - Ffriction).

Now this is exactly how much force in the direction up the ramp you need.
However, this force will be a component of the force in the horizontal direction (the direction the arm pushes).
that can be found by (Fg - Ffriction) / cos(27.5 degrees) where Fg and Ffriction are those from before (Fg is down the ramp, Ffriction is up the ramp [this is why it's negative].


The trickiest part of this problem is visualizing when to use sin and cos. The only angle in question is 27.5 degrees. Visualize an axis system where the positive y axis points out of the ramp and the positive x axis points along the ramp (down it, ideally). Gravity will still come down in your sketch, but in this coordinate system it can be broken into separate components.

I believe the solution is this ( (.405)(9.8)(6.25)(cos[27.5]) - (9.8)(6.25)(sin[27.5]) ) /cos27.5

This post was edited by ringo794 on Mar 9 2019 06:10pm
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