Static and Kinetic Friction Forces (Finding Net Force)

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The discussion centers on calculating the applied horizontal force required to initiate movement in a block with given coefficients of static and kinetic friction. The static friction force was correctly determined to be 21.3 N. For the net force after the block starts to slide, participants debated the correct approach, noting that the net force is not zero due to the transition from static to kinetic friction. It was clarified that the horizontal forces should be calculated by considering the kinetic friction force, which is lower than the static friction force. Ultimately, the correct value for the net force was identified as 14.19 N after resolving a calculation error.
FAJISTE
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Homework Statement



The coefficient of static friction between a block and a horizontal floor is 0.414, while the coefficient of kinetic friction is 0.138. The mass of the block is 5.24 kg. A horizontal force is applied to the block and slowly increased. What is the value of the applied horizontal force at the instant that the block starts to slide?

b) What is the net force on the block after it starts to slide?




Homework Equations



F=uN

The Attempt at a Solution



a)Okay, so the answer to the first part is 21.3 N.

F = uN where us is .414 and N is mg.

b) I don't get how to find the net force. So the mg and Normal force cancel out, so I calculated the horizontal forces using F=uN. I subtracted the static force from the kinetic force and the answer was wrong. What am I doing wrong?
 
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No information is given about the acceleration, which affects the answer. Some assumption must be made. You could assume the object is moving in uniform motion with no acceleration, in which case the net force is zero from F = ma. However, I suspect you are supposed to assume that the force you found in part (a) continues to be applied. Since the friction force is now reduced, the object will accelerate horizontally so there is a nonzero net force acting on it. Just add up the horizontal forces (not including F = ma which tells what acceleration the net force causes). The vertical forces add to zero as you said.
 
Yea what I'm not sure about is what forces they are.

Is it the static Friction force + the Kinetic Friction force? I really don't know what the horizontal forces are
 
bump sry but its due tomorrow
 
doesn't anyone know this?
 
bump bump bump
 
Please someone help me.
 
FAJISTE: For part (b), change the plus sign in post 3 to a minus sign.
 
Yeah i tried that too and the answer was wrong :(
 
  • #10
FAJISTE: You said you tried that, but what number did you try? Show your work so we can make sure we are talking about the same equation and numbers. How many tries are you allowed? I agree with the post by Delphi51. In other words, the only other logical tries for part (b) I can think of (so far), if the above fails (and show your work for that), are zero, and "unknown." Do they subtract points from you for multiple tries? Also, are you sure you did not omit any words from the given problem statement in post 1?
 
  • #11
n/m got it lol, i was putting 14.9 into the computer instead of 14.19

Thanks though!
 

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