Friction: what is the net force on the block the instant that it starts to slide

[paxian]

Homework Statement

The coefficient of static friction between a block and a horizontal floor is 0.40, while the coefficient of kinetic friction is 0.15. The mass of the block is 5.0 kg. If a horizontal force is slowly increased until it is barely enough to make the block start moving, what is the net force on the block the instant that it starts to slide?

Homework Equations

Fx= F-coefficient of static friction*m*g=0

The Attempt at a Solution

F= 0.4*5.0 kg*9.8m/s^2= 19.6N
However the answer on the key is 12N? I am not sure how to get to 12N? Please help... My thought was, when the block start moving, I have to use static coefficient. Your help is greatly appreciated!

 

[kjohnson]

If its moving you always use kinetic friction. Even if its barely sliding at constant velocity.

 

[1MileCrash]

You found the first step. However, that's just one force, not net force. There is another force, opposite, which is kinetic friction. Kinetic friction is there at the instant the block starts to slide, you found the applied force needed to get the block moving.

Find the force that kinetic friction will apply. Remember, it's in the opposite direction. IF you find the net force of what you found, the applied force, and kinetic friction, I think you can tell me why the answer is what it is.

 

[paxian]

You found the first step. However, that's just one force, not net force. There is another force, opposite, which is kinetic friction. Kinetic friction is there at the instant the block starts to slide, you found the applied force needed to get the block moving.

Find the force that kinetic friction will apply. Remember, it's in the opposite direction. IF you find the net force of what you found, the applied force, and kinetic friction, I think you can tell me why the answer is what it is.

Fx= Fstatic - Fkinetic = 0.4*5kg*9.8m/s^2-.15*5kg*9.8m/s^2=12.25N... I found the answer...but why do we have to subtract kinetic friction from static friction?

 

[1MileCrash]

We don't really do that.

While kinetic friction is found as a set force, static friction is found as a maximum force. Static friction balances out applied force until it's max is reached, at what point the applied force is greater than static friction and therefore starts moving.

So, by knowing that we "push the block with a force barely enough to move it" we simply use the maximum static friction to figure out the applied force.

We do not subtract kinetic friction from static friction. We use max static friction to know what force we needed to apply, then we subtract kinetic friction from that applied force.

 

[paxian]

We don't really do that.

While kinetic friction is found as a set force, static friction is found as a maximum force. Static friction balances out applied force until it's max is reached, at what point the applied force is greater than static friction and therefore starts moving.

So, by knowing that we "push the block with a force barely enough to move it" we simply use the maximum static friction to figure out the applied force.

We do not subtract kinetic friction from static friction. We use max static friction to know what force we needed to apply, then we subtract kinetic friction from that applied force.

I understand the logic behind it now! Thanks so much for your help!