# Questions on Forces

1. Aug 10, 2011

### fterh

Some background information: I'm doing some reading up for PhO (so it's beyond what I'm supposed to learn), and so I'll post all my questions here (regarding both concepts and actual practice questions). Sorry if you feel there is a lack of effort on my part, but sometimes I'm really lost and don't know where to start.

1) http://imgur.com/3Wyp7

My question: How come for the equality to hold, [ITEX]\mu_s \tan{\theta} > 1[/ITEX]? I can't get the link. And since we don't know what is N2 and N2 cannot be eliminated from the equation, how come there is no mention of N2 after the statement "For the equality to hold,"?

2. Aug 10, 2011

### bp_psy

First
$\ N_2(1-\mu_s \tan{\theta})+\frac{1}{2}mg=0$
then
$\ N_2(1-\mu_s \tan{\theta})<0$
therefore:
$\mu_s \tan{\theta} > 1$.
There is no mention of N2 since in the equation it N2 refers to the norm of the force and is a positive number.

3. Aug 12, 2011

### fterh

But how come $\ N_2(1-\mu_s \tan{\theta})<0$?

Shouldn't it be: $N_2(1-\mu_s \tan{\theta}) = -\frac{1}{2}mg$?

4. Aug 12, 2011

### cepheid

Staff Emeritus
I haven't read the problem, but it seems to me that the two quoted statements above are completely consistent with each other. Both are true.

5. Aug 20, 2011

### fterh

Can you explain how you arrived at that?

6. Aug 20, 2011

### cepheid

Staff Emeritus
If something is negative, then it is also < 0.

7. Aug 20, 2011

### kuruman

I will say the same thing that cepheid has been trying to point out differently.

We agree that $N_2(1-\mu_s \tan{\theta}) +\frac{1}{2}mg=0$
The second term in the above sum is positive. The only way that the two can add up to give zero is if the first term in the sum is negative.