# Friction on an incline

1. Oct 30, 2013

### Kot

1. The problem statement, all variables and given/known data
The hanging mass in the system is pulled down until the friction force is maximized between the other mass and the incline. Find the initial acceleration of the mass on the incline if the string is then cut. Let the average coefficient of kinetic friction be 80% the coefficient of static friction.

2. Relevant equations

$F=ma$ and $f=\mu N$

3. The attempt at a solution

So the problem states that the hanging mass is pulled until the friction force (static since it's at rest) is maximized. I drew free body diagrams for both masses here http://i.imgur.com/vHQ4vJy.jpg. I understand that the problem is asking me to find the maximum static friction of the block on the incline (before it starts to move) but I don't know how to express that in an equation. Could someone explain how I would do this?

*edit sorry for the big image, I'm not sure how to resize it.

Last edited: Oct 30, 2013
2. Oct 30, 2013

### cepheid

Staff Emeritus
Nice diagram!

So, according to your FBD1, sum of forces in the x direction yields:

T + fs = -Fs

So, what's countering the restoring force of the spring is the combination of the tension in the rope and static friction between the block and the plane. The problem is saying that the spring is extended until the maximum amount of static friction is developed. So, in other words this tells you that fs = μsN. When the string is cut, the tension is removed, and the force balance will be gone. There will be a net force on the block. Can you take it from here?

Last edited: Oct 30, 2013
3. Oct 30, 2013

### Kot

From what you said, the force of the spring is equal to the tension in the string plus the static friction of the block. I assume that this means the spring is stretched by a force $F=f_{s}+T$. Since the block is currently balanced, removing the Tension force would unbalance the forces on the x-axis. Then the net force on the block would act down the ramp until the force of the spring and the friction force is balanced?

4. Oct 30, 2013

### cepheid

Staff Emeritus
Yes

[post edited]

Yeah I guess. Initially, when a net force develops, it will be because Fs > fs (in magnitude). So, that is all the information you need to find the initial acceleration. Later on, once the block is actually moving, the friction will decrease because it will now be kinetic friction that takes over. The kinetic friction will be constant, since it depends only on N. The spring force depends only on how much it is stretched. Since the spring is becoming less stretched as the thing moves down the plane, I suppose eventually the spring force will decrease until it is equal to and then less than the kinetic friction, at which point the block will start to slow down.

5. Oct 30, 2013

### cepheid

Staff Emeritus
HMM! I just realized that my force balance equation missed a force from your FBD! There is a component of the weight that acts parallel to the inclined plane (which you have called the x-direction). We forgot gravity!

6. Oct 30, 2013

### Kot

Would I use F=ma, F being the force of the spring since Fs>fs. Then solve for a. Giving the initial acceleration a = Fs/m ?

*edit
I also forgot to mention that I decomposed the weight into x and y components. But that isn't used in my solution so I assume I did something wrong :(

7. Oct 31, 2013

### cepheid

Staff Emeritus
Yes, you'd use Fnet = ma, but Fnet would not be Fs. Fnet would be the net NET force in the x-direction. So you, have to compute that. Hint: the net force is what's left over after you add everything up, taking directions into account. ƩFx = ma. Initially, this sum yields zero and everything is balanced. Afterwards, the tension is removed, and this sum becomes non zero. Hence, 'a' becomes non-zero.

Your force balance equation has to include ALL of the forces in the x-direction. So, in addition to the tension, spring force, and static friction, you should also have the x-component of the weight in the sum.

Last edited: Oct 31, 2013