Quote:
Originally Posted by John
F = sqrt(36 + F2^2)
F = 10*(.2 + .6 (theta/(pi/2)))
theta = arctan(6/F2)
F = 10*(.2 + .6 (2arctan(6/F2)/pi))
sqrt(36 + F2^2) = 2 + 6 (2arctan(6/F2)/pi)
F2 ~= 2.491 N
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Excellent.
Quote:
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I'm confused; does this mean that it takes MORE sideways force to move the block than it would if there was no vertical force? For example, wouldn't the block move if F1 = 0N and F2 = 2N?
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Excellent. You have asked the $64K question.
If you
assume that the coefficient in any given direction is
not increased by force components not in that direction, then the answer is that the block will start to slip when F
2=2 Newtons. See attached "solution.pdf".
However, these 2 assumptions:
1: linear change in mu for angles between 0 and pi/2, and
2: mu is not increased by force components not in direction being considered
are I think open questions.
In FRC, there are wheels which are said to have different mu in the forward and sideways directions. I have looked but never seen any test data to show what happens when force is applied at a variety of angles between 0 and pi/2.