# Need some help with a trigonometry problem **URL:** \ **Category:** Factual Questions\ **Created:** [June 1, 2022, 9:41pm UTC](https://boards.straightdope.com/t/need-some-help-with-a-trigonometry-problem/965526 "2022-06-01T21:41:55Z")\ **Posts on this page:** 5\ **Page:** 6
**Author:** ![Robot\_Arm](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robot_arm/32/18280_2.png) [@Robot\_Arm](https://boards.straightdope.com/u/Robot_Arm)\ **Post date:** [July 16, 2024, 7:03am UTC](https://boards.straightdope.com/t/need-some-help-with-a-trigonometry-problem/965526/101 "2024-07-16T07:03:34Z")
> [@Snarky\_Kong](#): > > A civil engineer generally would have a full semester course on that sort of thing (statics). Kinda depressing that it’s taken me two years to get from chapter 4 to chapter 8. ---
**Author:** ![Manlob](https://avatars.discourse-cdn.com/v4/letter/m/96bed5/32.png) [@Manlob](https://boards.straightdope.com/u/Manlob)\ **Post date:** [July 16, 2024, 2:57pm UTC](https://boards.straightdope.com/t/need-some-help-with-a-trigonometry-problem/965526/102 "2024-07-16T14:57:43Z")
Much of the trig can be avoided, at the expense of more effort in figuring out the various angles in the diagram, by working with the force vector, R, acting at point B, rather than its components normal and tangent to the surface (N\_B and F\_B). Force balance along direction of R immediately gives magnitude of R: R=W cos(alpha+phi) since the force at A has no component in the R direction. Moments about point A gives 2hR sin(theta-phi) = hW sin(alpha+theta). Substituing for R results in 2 cos(alpha+phi) sin(theta-phi) = sin(alpha+theta) This is the point you had almost reached after all the trig work above. Using the identity for 2cos(x)sin(y) results in sin(theta-alpha-2 phi)=0 theta=alpha+2 phi, and if friction acts in the direction opposite from assumed, use negative phi, for theta=alpha-2 phi ---
**Author:** ![Robot\_Arm](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robot_arm/32/18280_2.png) [@Robot\_Arm](https://boards.straightdope.com/u/Robot_Arm)\ **Post date:** [July 19, 2024, 4:27am UTC](https://boards.straightdope.com/t/need-some-help-with-a-trigonometry-problem/965526/103 "2024-07-19T04:27:16Z")
> [@Manlob](#): > > \begin{equation} 0=−sin(α+θ)+2(−sinϕsinα+cosϕcosα)(cosϕsinθ−sinϕcosθ) \end{equation} I think you might be missing a term. I get: \begin{equation} \phi=tan^{-1}\mu\_{s}\\ tan \phi=\mu\_{s}\\ \end{equation} \begin{equation} 1+\mu\_{s}^2\\ =1+\tan^2\phi\\ =\sec^s\phi\\ \end{equation} \begin{equation} -W\sin\lgroup\alpha+\theta\rgroup+\frac{2W\lgroup-\mu\_{s}\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\mu\_{s}\cos\theta\rgroup}{1+\mu\_{s}^2}=0\\ -W\sin\lgroup\alpha+\theta\rgroup+\frac{2W\lgroup-\mu\_{s}\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\mu\_{s}\cos\theta\rgroup}{\sec^2\phi}=0\\ -W\sin\lgroup\alpha+\theta\rgroup+2W\lgroup-\mu\_{s}\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\mu\_{s}\cos\theta\rgroup\cos^2\phi=0\\ -W\sin\lgroup\alpha+\theta\rgroup+2W\lgroup-\tan\phi\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\tan\phi\cos\theta\rgroup\cos^2\phi=0\\ -W\sin\lgroup\alpha+\theta\rgroup+2W\cos\phi\lgroup-\sin\phi\sin\alpha+\cos\phi\cos\alpha\rgroup\lgroup\cos\phi\sin\theta-\sin\phi\cos\theta\rgroup\\ \end{equation} > [@Manlob](#): > > Much of the trig can be avoided, Eh, in for a penny. I’ve come this far, I would like to see if I can get this to resolve to something usable. ETA: Forgot to factor out W. \begin{equation} -\sin\lgroup\alpha+\theta\rgroup+2\cos\phi\lgroup-\sin\phi\sin\alpha+\cos\phi\cos\alpha\rgroup\lgroup\sin\phi\sin\theta-\sin\phi\cos\theta\rgroup\\ \end{equation} ---
**Author:** ![Manlob](https://avatars.discourse-cdn.com/v4/letter/m/96bed5/32.png) [@Manlob](https://boards.straightdope.com/u/Manlob)\ **Post date:** [July 23, 2024, 2:49am UTC](https://boards.straightdope.com/t/need-some-help-with-a-trigonometry-problem/965526/104 "2024-07-23T02:49:18Z")
> [@Robot\_Arm](#): > > I think you might be missing a term. I get: You have an extra cos(phi). After you multiplied each of the 2 terms in parentheses by cos(phi), you kept an extra factor of cos(phi) in the next line. ---
**Author:** ![Robot\_Arm](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robot_arm/32/18280_2.png) [@Robot\_Arm](https://boards.straightdope.com/u/Robot_Arm)\ **Post date:** [July 23, 2024, 3:45am UTC](https://boards.straightdope.com/t/need-some-help-with-a-trigonometry-problem/965526/105 "2024-07-23T03:45:13Z")
> [@Manlob](#): > > You have an extra cos(phi). That’s why I posted; I feel like you left the cos(phi) term out. Look at the line before that. \begin{equation} -W\sin\lgroup\alpha+\theta\rgroup+2W\lgroup-\tan\phi\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\tan\phi\cos\theta\rgroup\cos^2\phi=0\\ \end{equation} The last term is cos^2(phi). Multiply the preceding terms by cos(phi): \begin{equation} \lgroup-\tan\phi\sin\alpha+\cos\alpha\rgroup\lgroup\sin\theta-\tan\phi\cos\theta\rgroup\cos\phi\\ \lgroup-\sin\phi\sin\alpha+\cos\phi\cos\alpha\rgroup\lgroup\cos\phi\sin\theta-\sin\phi\cos\theta\rgroup \end{equation} and there’s still another cos(phi) term left, isn’t there? [Previous page](https://boards.straightdope.com/t/need-some-help-with-a-trigonometry-problem/965526.md?page=5)