
#1
Mar1212, 03:51 PM

P: 11

1. The problem statement, all variables and given/known data
A 45kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 24 degrees , the crate begins to slide downward. What is the coefficient of static friction between the crate and the ramp? 2. Relevant equations [itex]\Sigma[/itex]F=ma fs=[itex]\mu[/itex]sN 3. The attempt at a solution I started by drawing a free body diagram. My weight came out to be 441 N. I have the angle of my weight at 24 degrees. The normal force at mgcos24, for which I got 187N. mgsin24 came out to be 399N which is also the static friction force. Then I just substituted into the static friction equation and my coefficient of static friction came out to be 2.13. I entered this into the homework problem which we do online and it is wrong. Did I miss a step? What did I do wrong? 



#2
Mar1212, 04:01 PM

P: 157

Your values for mgcos24 and mgsin24 are incorrect. Your calculator appears to be in radian mode.




#3
Mar1212, 04:06 PM

P: 297

The applied formula is right.The math is wrong.
the weight u calculated using m as 450 and g as 9.8 (all in SI units) is right as 441 N. As u said normal force is mgcos24. and mgsin24 as component of weight parallel to inclined plane. However, the calculated values are not right. Hint(for angles less than 45 degrees cos x is always greater the sin x. However, here your mgcos24 is less than mgsin24) 



#4
Mar1212, 04:08 PM

P: 11

Can someone help me with this static friction problem?
Wow, thanks a lot Calc class, lol.
Got the answer, thanks for your help guys 


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