forces diagram: A particle, q, of mas m kg, is in equilibrium on a smooth plane...

[petr]

Hello,

I am stuck at this question:

A particle, q, of mas m kg, is in equilibrium on a smooth plane which makes an angle of 60 to the vertical. This is achieved by an attached string S, with tension 70N, angled at 10 to the plane, as shown. Draw a force diagram and find both the mass of q and the reaction on it from the surface.

here s the picture of the given picture.
https://onedrive.live.com/redir?res...authkey=!AGJ8evsZQ8ss8CE&v=3&ithint=photo,jpg
question 5)

and here is the official correct solution for this question.
https://onedrive.live.com/redir?res...authkey=!AHM2-WAn-TypBSs&v=3&ithint=photo,jpg

and the problem is I don't understand how they figure out the angles in the triangle. i m working on in by the morning and it drives me crazy.
https://onedrive.live.com/redir?res...authkey=!AAqZ01x6vOGDT5s&v=3&ithint=photo,jpg

the question would not make me any problem if I knew how to figure out the angles in there.

Thank you very much for any sort of help.

 

[stapel]

To the original poster: In general, it is more helpful to provide the information here, rather than making people follow multiple links and copy the information out for you.

For other viewers: The intended text, as near as can be determined, appears to be as follows:

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5. A particle, Q, of mass m kg, is in equilibrium on a smooth plane which makes an angle of 60 degrees to the vertical. This is achieved by an attached string, S, with tension 70 N, angled at 10 degrees to the plane, as shown:

. . . . .

Draw a force diagram and find both the mass of Q and the reaction to it from the surface.

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Huge hint: The angle of the plane to the horizontal (in this case, 30 degrees) will always be the angle indicated in the image:

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\(\displaystyle \begin{align} \mbox{Sine rule: }\,\dfrac{mg}{\sin(100^{\circ})}\, &=\, \dfrac{70}{\sin(30^{\circ})} \\ \\ \mbox{so: }\, mg\, &=\, \dfrac{70\, \sin(100^{\circ})}{\sin(30^{\circ})} \\ \\ m\, &=\, 14.1\, \mbox{ kg} \\ \\ \dfrac{R}{\sin(50^{\circ})}\, &=\, \dfrac{70}{\sin(30^{\circ})} \\ \\ \mbox{so: }\, R\, &=\, \dfrac{70\, \sin(50^{\circ})}{\sin(30^{\circ})}\, =\, 107\, \mbox{ N} \end{align}\)

Yet another example of how triangles might just save your mark for the M1 module. Although you could also have solved it by resolving forces parallel and perpendicular to the slope if that floats your boat.

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The following is the original poster's work: