# What's affecting the friction?

1. Jul 2, 2013

### null void

I am quite confused when my lecturer told me that the mass of an object doesn't affect the static friction.

F = μN , the N suppose to be the Normal force right? If i have an object on an horizon surface, the N should be -mg, why doesn't mass affect the static friction?

2. Jul 2, 2013

### Staff: Mentor

Note that for static friction, μN gives the maximum value of the friction force. The actual friction force will be equal or less than that. So you need to be clearer about the context of your question.

Example: A block rests on a table. You push with a horizontal force of 1 N and it doesn't move. What's the value of the static friction force? What if the mass of the block doubled? Will that change the static friction?

3. Jul 2, 2013

### torquil

For static friction, the friction force is of the same magnitude as the force with which you are pushing on the object, but oppositely oriented, so as to exactly keep the object stationary. Since we are talking about static friction, we are assuming that you are pushing with a force not larger in magnitude than $\mu N$. This upper limit depends on the mass, through N.

4. Jul 3, 2013

### null void

I was doing an experiment by putting a wooden block on a plane, then slowly lifting up the other side to make the plane incline more until the wooden block slip( when the force that the friction is opposing exceed the static friction). I know that when lifting up the plane, the N force getting smaller then the force perpendicular to the N will become bigger.

the static force = 1 N; when the mass doubled, i suppose the friction also doubled. So in conclusion the static friction does affected by mass?

5. Jul 3, 2013

### kulkajinkya

Q: A block rests on a table. You push with a horizontal force of 1 N and it doesn't move. What's the value of the static friction force? What if the mass of the block doubled? Will that change the static friction?
Ans: Value is 1N. Even if the mass is doubled, the force remains 1N, just the maximum allowable static friction becomes double.
So originally if the block would move at 10N(say), it will now not move until 20N is applied.
I hope this helps

6. Jul 3, 2013

### Staff: Mentor

OK. Since you're finding the angle at which the block just begins to slip, the static friction at that point will equal μN, the maximum value. As you point out, the greater the angle of the incline, the smaller the value of that normal force.

In any case, should the angle at which the block just begins to slip depend on the mass of the block?

Right.

Why do you think that? You're still pushing with only 1 N of force. (What changes is the maximum value of the friction force, not the actual friction force.)

More examples. Two blocks on a horizontal table. One with a mass of 1 Kg and the other with a mass of 10 Kg. Let's say that μ = 0.5. So what is the maximum value of static friction in each case, which is given by μN = μmg?

For the 1 kg block: μmg = 4.9 N
For the 10 kg block: μmg = 49 N

So, if you push the 1 kg block with a horizontal force of 1 N, what will be the static friction created? (Hint: How does the 1 N force compare to the maximum value for the static friction?)

Same question for the 10 kg block.

See above.

7. Jul 3, 2013

### CWatters

At the instant the block slips the force down the slope is equal to the max value of static friction.

Write equations for both and equate them.

You should find that the mass cancels.

You can then rearrange the result to solve for the coefficient of static friction which will be independent of mass.

That's probably what your lecturer meant.

8. Jul 6, 2013

### null void

I get it, thanks