This is a great example of how to calculate forces for a static (no motion) system, and then what happens if you break loose and allow motion to happen. Note how the coordinate system was oriented to make the math a lot easier.
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Yes, that does! Thank you!
It actually does matter. I should have drawn in my coordinate system so you could see it more clearly. X is along the “T” force and y-axis is along the “Fn” force.
The problem gave theta = 27 degrees so we know that from the original image.
If you think about the ramp being flat along the ground, no incline at all, notice how the “mg” and “Fn” forces are parallel. What happens if you lift the right edge of the ramp just a tiny bit? What happens to “mg” and “Fn”?
“mg” doesn’t change, but “Fn” does, and now they are no longer parallel by that “tiny” bit of an angle that you raised the platform.
The more we raise the right side, the more that angle is going to change.
From trigonometry, it’s called “alternating interior angles”.
Does that help?
At 1:17, you find an angle and declare it to be theta. How did you do that? I can’t seem to understand how you can just take an angle from one direction and rotate it to another direction and know that it is in the right position.
For the cut rope part (around 5:30 in the video), you’ll find gravity in the FBD as “mg”. It looks like I had another typo when I wrote the equation from the FBD but had the correct answer. I’ve fixed the video so you can view it now.
A couple of questions here, on the last problem (the “cut rope”):
– why doesn’t gravity come into play? Seems like it’d be impossible to solve this without considering gravity. For example, in space, the block wouldn’t accelerate at all when the rope was cut.
– why is the calculation based on sin (theta)? The use of sin and cos in previous problems up to this point has pretty much made sense to us, but here, it seems like sin (theta) would give a value for the “magnitude” of a straight horizontal line, which would not align to the chosen coordinate system.