0
$\begingroup$

I am a junior level mechanical engineering student, and have taken physics, statics, dynamics, etc. so I know how to do this problem, though something seems to be tripping me up.

Given a standard ramp friction problem such as the one below, and given a coefficient of static friction $\mu_s$, and knowing it is not moving, we can create a system of equations to determine all forces in terms of the mass. $$\vec{F_g} + \vec{F_N} + \vec{F_f} = \vec{0}$$ If we define $\hat{x}$, $\hat{y}$ to be unit vectors parallel, perpendicular to the bottom surface of the ramp: $$-mg \hat{y} + \sin{\theta} F_N \hat{x} + \cos{\theta} F_N \hat{y} - \cos{\theta} F_f \hat{x} + \sin{\theta} F_f \hat{y} = 0 \hat{x} + 0 \hat{y}$$ By substituting $F_f = \mu_k F_N$, and seperating the above into an equation solely with the $\hat{y}$ components, we obtain the following: $$F_N = \frac{mg}{\mu_s \sin(\theta) + \cos(\theta)}$$

If we define $\hat{i}$, $\hat{j}$ to be unit vectors parallel, perpendicular to the top surface of the ramp, respectively: $$F_N \hat{j} - F_g \cos{\theta} \hat{j} = 0 \hat{j}$$ therefore, $$F_N = mg \cos{\theta}\,.$$

Selecting arbitrary values for $\theta$ and $\mu_s$ shows that these two expressions are not equal. It is obvious that a coordinate rotation should not change our answer, so what is happening?

Standard Ramp Problem

$\endgroup$
1
  • $\begingroup$ Both equations are correct, it's just that they determine both $N$ and $\theta$. $\endgroup$
    – march
    Commented Mar 10, 2023 at 20:44

2 Answers 2

0
$\begingroup$

Both equations are correct, it's just that they determine both $N$ and $\theta$, so you've got two equations that you can then solve for these two quantities. In other words, there is only a particular angle at which all of these things can be true at once.

$\endgroup$
1
  • $\begingroup$ Kind of forgot about this question, but this was it. I was thinking of this as a statics problem, making the incorrect assumption that it wasn't moving, when in fact, it was. $\endgroup$ Commented Mar 19, 2023 at 20:47
0
$\begingroup$

Because the friction is static, your equation for friction is only the maximum possible static friction. There is no formula for static friction. In your first example (i.e., horizontal & vertical), leave $F_f$ as $F_f$. You must use both horizontal and vertical components to eliminate the unknown $F_f$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.