## static friction and kinetic friction, solve for acceleration and force.

1. The problem statement, all variables and given/known data

A 62 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 16° above the horizontal. (a) If the coefficient of static friction is 0.57, what minimum force magnitude is required from the rope to start the crate moving? (b) If μk = 0.31, what is the magnitude of the initial acceleration (m/s^2) of the crate?

2. Relevant equations

3. The attempt at a solution
Attempt of number (a)
fs=Fn*Ms
Let Fn + Tsin15=mg
Fn = mg - Tsin15

Attempt of number (b)
ma= Tcos15 - fk (Do I need to subtract fs in this equation?)

fs=magnitude of static friction. In this case, it's the maximum value of static friction.
Fn=normal force
T= tension of the rope
fk=magnitude of kinetic friction.

Are the equations of my attempts correct ?
How do find tension ? or is there anyway to cancel it out ?

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 Double checked my question. No mistake.
 Apparently, I made some mistakes on the equations above. On the question, it states the minimum force to move the crate. As I can imagine, at the very small instant, "the bond is broke". Therefore, for (a), the equation would look like this T cos 16 - fs,max = 0 (the minimum force to " break the bond") solve for tension. (fs,max= maximum value of static friction) Then, substitute back into the equation of the (b), to solve for acceleration. If I make any mistake, pls correct me. Thank you.

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## static friction and kinetic friction, solve for acceleration and force.

 Quote by tebes 1. The problem statement, all variables and given/known data A 62 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 16° above the horizontal. (a) If the coefficient of static friction is 0.57, what minimum force magnitude is required from the rope to start the crate moving? (b) If μk = 0.31, what is the magnitude of the initial acceleration (m/s^2) of the crate? 2. Relevant equations 3. The attempt at a solution Attempt of number (a) fs=Fn*Ms Let Fn + Tsin15=mg Fn = mg - Tsin15
This is the correct equation for Fn. But now you need an equation in the x direction to solve for T
 Attempt of number (b) ma= Tcos15 - fk
correct
 (Do I need to subtract fs in this equation?)
Once the crate starts moving, static friction no longer applies
 fs=magnitude of static friction. In this case, it's the maximum value of static friction. Fn=normal force T= tension of the rope fk=magnitude of kinetic friction. Are the equations of my attempts correct ? How do find tension ? or is there anyway to cancel it out ?
Find it in part 'a'

 Quote by PhanthomJay This is the correct equation for Fn. But now you need an equation in the x direction to solve for T correct
(Do I need to subtract fs in this equation?)[/quote]Once the crate starts moving, static friction no longer applies Find it in part 'a'[/QUOTE]

Thank you. One more question, if a block is at rest, then start to move, does the force need to overcome the static friction ?

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 Quote by tebes (Do I need to subtract fs in this equation?)
Once the crate starts moving, static friction no longer applies Find it in part 'a'
 Thank you. One more question, if a block is at rest, then start to move, does the force need to overcome the static friction ?
At the max value of static friction, the block is on the verge of moving, that is, motion is imminent under the applied force . Now when you increase the tension theoretically just an infinitesimal amount above that force, the block starts to move and the kinetic friction force takes over from the staic friction force, and the crate accelerates until you reduce the tension to keep it moving at constant speed, if you so desire.

 Quote by PhanthomJay Once the crate starts moving, static friction no longer applies Find it in part 'a' At the max value of static friction, the block is on the verge of moving, that is, motion is imminent under the applied force . Now when you increase the tension theoretically just an infinitesimal amount above that force, the block starts to move and the kinetic friction force takes over from the staic friction force, and the crate accelerates until you reduce the tension to keep it moving at constant speed, if you so desire.

Thank you.

 Tags acceleration, force, friction, kinetic coefficient, static