Friction force acting on a block
I have attached a picture file and a free-body diagram file for this problem.
Determine the friction force acting on the block shown when the angle is 30 degrees and P=200N.
us=0.3 (static friction)
uk=0.2 (kinetic friction)
The free-body diagram is how I started this solution. My equations became:
Fx=0=200-Ncos(60)-fcos(30)
Fy=0=-1000-fsin(30)+Nsin(60)
When I tried to substitute in uN for f I got two different answers for N. I tried this method for usN and ukN. Where am I going wrong?
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If you assume in advance that the block will not move, then the equations you wrote down are correct. But they are not written down in a convenient coordinate system. Let's work with your equations first.
The assumption that the block will not move can be incorrect. However, this is not a problem, because you can use your equations to solve for both the normal force N and the friction force f. The value for f you get must then be interpreted as the friction force that is necessary for the block to remain at rest. You then calculate the ratio of f and N to see if it exceeds us. If not, then the block will indeed remain at rest. If the ratio exceeds us, then the block will accelerate. The friction force of f you ...
Solution Summary
The friction force acting on a block is determined. A free-body diagram is provided for the question.