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Problem: Determine the least angle $\theta$ where equilibrium is possible. Express $\tan{\theta}$ in terms of $\alpha$ and $\mu_s$

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Attempt: I started of with drawing out all the acting forces:

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I know that $\tan{\theta}=\frac{h}{x_1+x_2}$ so I need to express $h$ and $x_1,x_2$ in terms of $\alpha$ and $\mu_s.$ I know the following:

$$x_1+x_2 = l\cos{\theta} \\ h=l\sin{\theta} \\ x_1=h\tan{\left(\alpha-\frac{\pi}{2}\right)} \\ x_2 = l\cos{\theta}-h\tan{\left(\alpha-\frac{\pi}{2}\right)}$$

and for the forces:

$$N_{Ax}=N_A\sin{\alpha} \\ N_{Ay}=N_A\cos{\alpha} \\ F_f=N_B\mu_s$$

$\sum F_x = 0 = N_{Ax}-F_f = N_A\sin{\alpha}-N_B \mu_s \Rightarrow N_A = \frac{N_B\mu_s}{\sin{alpha}}.$

$\sum F_y = 0 = N_{Ay} - mg + N_B = N_A\cos{\alpha}-mg-N_B \Rightarrow N_A=\frac{mg-N_B}{\cos{\alpha}}.$

Setting the above equal gives

$$\left\{ \begin{array}{rcr} N_A & = & \frac{mg\mu_s}{\mu_s\cos{\alpha}+\sin{\alpha}} \\ N_B & = & \frac{mg\sin{\alpha}}{mu_s\cos{\alpha}+\sin{alpha}} \\ \end{array} \right.$$

Aand now I don't really know what to do with this information.

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closed as off-topic by Emilio Pisanty, Gert, stafusa, John Rennie, Jon Custer Dec 3 '17 at 20:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, Gert, stafusa, John Rennie, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Please read our guidelines for homework and exercise questions carefully. Note that showing your on workings is not sufficient; you need to ask about the underlying concepts that you're confused about. $\endgroup$ – Emilio Pisanty Dec 2 '17 at 18:14
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    $\begingroup$ I thought I was clear about needing to express $h$ and $x_1,x_2$ in terms of $\alpha$ and $\mu_s.$ And by the way, who told you this is a homeworkquestion? I'm simply studying for my exam after Christmas, We don't have "homework" in universities here in my country, we simply study for an exam. Quite annoying that one can't even ask a question about physics in a physics forum without having to go through a hassle with some bureaucratic moderator. $\endgroup$ – Parseval Dec 2 '17 at 18:19
  • $\begingroup$ @Parseval: if we didn't put the 'brakes' on homework style questions, we'd be inundated with them. As it stands we still are. Nothing to do with 'bureaucracy', just common sense. Voted to close. Other forums don't have the same policy, so try there? $\endgroup$ – Gert Dec 2 '17 at 19:11
  • $\begingroup$ This is still not a homework. And I've made it clear where I'm stuck. I don't see the problem? $\endgroup$ – Parseval Dec 2 '17 at 19:17
  • $\begingroup$ Explained by Emilio. We call them homework style questions. It doesn't literally by homework. Check the policy. $\endgroup$ – Gert Dec 2 '17 at 20:21