- #1
ShizukaSm
- 85
- 0
Find the friction coefficient between the bar and the surfaces. Both surfaces are equal (and thus have equal coefficients).
Ok so, my problem is that, generally in physics when I assume an wrong direction for a vector, I will be able to discover that in my answer (generally speaking the vector will be negative), however, in this question assuming a wrong direction will give me a completely wrong answer and I don't understand why this happens or how to define the right orientation right from the start.
My original trial:
[itex]\\\sum F_x = 0 \rightarrow N_a\mu_e + Nb\left(-\mu_e \cos(\theta)-\sin(\theta) \right ) = 0
\\\sum F_y = 0 \rightarrow N_a + N_b(\cos(\theta)-\mu_e \sin(\theta)) = W
\\\sum \tau_A = 0 \rightarrow W = \frac{N_b}{\sin(\theta)}
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\mu_e = 0.569345 \text{ or } \mu_e = -1.75641[/itex]
Book (I suppose it's also the correct one):
[itex]\\\sum F_x = 0 \rightarrow N_a\mu_e + Nb\left(+^{[1]}\mu_e \cos(\theta)-\sin(\theta) \right ) = 0
\\\sum F_y = 0 \rightarrow N_a + N_b(\cos(\theta)+^{[2]}\mu_e \sin(\theta)) = W
\\\sum \tau_A = 0 \rightarrow W = \frac{N_b}{\sin(\theta)}
\\\text{Only differences in [1] and [2].}
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\mu_e = 0.225979 \text{ or } \mu_e = 4.42518[/itex]