# Questions on Forces: How Come \mu_s \tan{\theta} > 1?

• fterh
In summary: If something is negative, then it is also < 0.But how come \ N_2(1-\mu_s \tan{\theta})<0 ?It could be because the second term in the sum is positive.
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,"?

fterh said:
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,"?
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.

bp_psy said:
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.

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

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

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

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

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.

cepheid said:
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.

Can you explain how you arrived at that?

fterh said:
Can you explain how you arrived at that?

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

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

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

## 1. What is the meaning of \mu_s and \tan{\theta}?

\mu_s is the coefficient of static friction, which is a measure of the force needed to start an object moving on a surface. \tan{\theta} represents the tangent of the angle of inclination between the surface and the object.

## 2. Why is it important that \mu_s \tan{\theta} is greater than 1?

When \mu_s \tan{\theta} is greater than 1, it means that the force needed to start an object moving is greater than the force of gravity pulling it down the incline. This is important because it determines whether an object will be able to stay in place or slide down the incline.

## 3. How do \mu_s and \tan{\theta} affect each other?

The value of \mu_s depends on the materials of the object and the surface it is on, while \tan{\theta} depends on the angle of inclination. As \tan{\theta} increases, the force required to start the object moving also increases. Similarly, as \mu_s increases, the force needed to start the object moving also increases.

## 4. What happens when \mu_s \tan{\theta} is less than 1?

When \mu_s \tan{\theta} is less than 1, it means that the force needed to start an object moving is less than the force of gravity pulling it down the incline. In this case, the object will slide down the incline.

## 5. Can \mu_s \tan{\theta} be equal to 1?

Yes, \mu_s \tan{\theta} can be equal to 1. This means that the force required to start an object moving is equal to the force of gravity pulling it down the incline. In this case, the object will be on the verge of sliding, but will not actually move.

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