# I solved a problem but I need some insight into why it was solved this way.

1. Jan 9, 2006

### G01

Hi. I just solved a homework problem. I eventually got the right answer and am only confused about why it was solved that way. This is the reason I posted it here instead of in the homework help section. All I need is a little insight into why the solution is the way it is. OK here we go:

A 1kg block sits atop a 5kg block as it osscilates on a spring. The period is 1.5s. The small block begins to slip when the Amplitude is increased to .4m What is the coefficient of satic friction bewteen the two blocks.

The solution is simple:

$$F_{sp} = f_s$$

$$-kA = \mu_smg$$

$$\mu_s = \frac{-kA}{mg}$$

Using the period you can solve for k = 105.3 N/m. The you can solve for $$\mu_s$$

Now heres my problem. In order to get the right answer you must use m= the total mass of the system= 6kg. I thought that since only the small block was sliding, only its mass shouldbe used in this calculation. Why is the total mass used? That is my question. Thank you for your help. (If this should still be in the homework help section, please move it.)

2. Jan 9, 2006

3. Jan 9, 2006

### Staff: Mentor

While your answer happens to be correct, the solution is not. You seem to be assuming that the force of the spring equals the static friction force. Why would that be?

The way to solve it is this: The only force on the small mass is the static friction. What's the maximum possible static friction? Thus what's the maximum allowable acceleration of that small mass before it begins to slide? Set that equal to the acceleration of the whole thing when the spring force is at its maximum.

See my comments above. Note that you are solving for the coefficient that will allow both masses to move together without slipping. The entire thing oscillates.

4. Jan 9, 2006

### G01

Ahh i see. so what your saying is this. The maximum acceleration from friction is

$$\mu_sg = a$$

now I set that equal to the acceleration of the whole system which would be:

$$\frac{-kA}{m_{system}} = a = \mu_sg$$

$$\frac{-kA}{m_{system}g} = \mu_s$$

Now that would give me the right answer, and now my reasoning is correct I hope. I understand this much better now I think. Thanks alot.

5. Jan 9, 2006

### Staff: Mentor

You got it. But watch out for that minus sign, lest someone roll their eyes at you like this: