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Physics Quiz 4
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Figure1 is a top view of a block sitting on a table. The block weighs 10 Newtons. The coefficient of static friction between the block and the table is 0.8 (the same in all directions). A force F1 = 6 Newtons is applied to the block (yellow arrow). A force F2 (blue arrow) is then slowly applied in the direction shown. At what magnitude of F2 does the block begin to slide? |
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2*SQRT(7) Newtons
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Thank you. |
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F2 >= 5N and some change.
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Make the following changes to the original problem: - The static coefficient of friction in the "Y" direction (the direction of the yellow arrow) is mu_y = 0.8 - The static coefficient of friction in the "X" direction (the direction of the blue arrow) is mu_x = 0.2 - The static coefficient of friction varies linearly with theta from mu=0.2 in the "X" direction to mu=0.8 in the "Y" direction as theta goes from zero (X direction) to pi/2 (Y direction). Now find the value of Fx at which the block begins to slide. |
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Fix that, get a new numerical value for F2, then ask yourself "what happens when F2 = 2 Newtons?" |
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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 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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Jason |
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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 F2=2 Newtons. See attached "solution.pdf". However, these 2 assumptions: 1: linear change in mu for angles between 0 and pi/2, andare 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. |
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Jason |
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