Static friction and kinetic friction, solve for acceleration and force.

In summary: One more question, if a block is at rest, then start to move, does the force need to overcome the static friction ?Yes, the force must overcome the static friction in order for the block to start moving. This is because static friction acts to resist the motion of an object at rest. So in order for the object to overcome this resistance and start moving, the applied force must be greater than the maximum static friction force.
  • #1
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Homework Statement



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?

Homework Equations





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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  • #2
Double checked my question. No mistake.
 
  • #3
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.
 
  • #4
tebes said:

Homework Statement



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?

Homework Equations





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'
 
  • #5
PhanthomJay said:
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 ?
 
  • #6
tebes said:
(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.
 
  • #7
PhanthomJay said:
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.
 

1. What is the difference between static friction and kinetic friction?

Static friction is the force that prevents two surfaces from sliding past each other when there is no motion. It is typically greater than kinetic friction, which is the force that opposes the motion of two surfaces when they are in contact and sliding past each other.

2. How do you calculate the acceleration of an object using static friction and kinetic friction?

To calculate the acceleration of an object using static friction and kinetic friction, you need to know the mass of the object, the coefficient of static friction, and the coefficient of kinetic friction. The acceleration can be calculated using the formula a = μs/m for static friction and a = μk/m for kinetic friction, where μs is the coefficient of static friction, μk is the coefficient of kinetic friction, and m is the mass of the object.

3. How does the force of static friction change with respect to the applied force?

The force of static friction increases as the applied force increases, up to a certain point where it reaches its maximum value and the object starts to move. Once the object is in motion, the force of static friction is replaced by the force of kinetic friction, which remains constant regardless of the applied force.

4. Can the coefficient of kinetic friction ever be greater than the coefficient of static friction?

No, the coefficient of kinetic friction is always less than or equal to the coefficient of static friction. This is because the force of static friction is typically greater than the force of kinetic friction, and the coefficient is a ratio of these two forces.

5. How do you solve for the force of static friction and kinetic friction in a given scenario?

To solve for the force of static friction and kinetic friction, you need to know the mass of the object, the coefficient of static friction, and the coefficient of kinetic friction. The force of static friction can be calculated using the formula Fs = μsN, where N is the normal force between the two surfaces. The force of kinetic friction can be calculated using the formula Fk = μkN. The normal force can be calculated using the formula N = mg, where m is the mass of the object and g is the acceleration due to gravity.

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