Friction Problems

  • Thread starter 404
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  • #1
404
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The coefficient of static friction between the 3.00 kg crate and the 35 degree incline is 0.300. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

I managed to do most of the question(whether it's wrong or right), Figured out friction was 7.23N, and the box slides down with 9.64 N if the force is not applied... But what I'm confused on is, how do you use a Y component to cancel out the x component? (I mean if you draw a coordinate plane perpendicular to the incline, then the 9.64 is going straight down the x-axis and the perpendicular froce being asked is going straight down the y axis)

Truck (10,000 kg) carries a load (20,000 kg), moving at 12 m/s. If load is not tied down and coefficient of static friction is 0.500 with truck bed, calculate the minimum stopping distance for which the load will not slide forward relative to the truck. And is there any unless info?

I figure if the load can't slide forward, the net force should be 0, therefore accerlation must be 0 too, but that's where I'm stuck...
 

Answers and Replies

  • #2
zwtipp05
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Force of friction is mu times the normal force. So if you increase the normal force, what happens to the force of friction?
 
  • #3
404
52
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ahh right friction increases too, so that's how you cancel out the x component of gravity... Thanks, I'll go try it now.
 

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