I am curious if I am setting this problem up right.

Three forces act on the bracket. Determine the magnitdue and direction theta of F2 so that the resultant force is directed along the positive u axis and has a magnitude of 50lb.

[URL=http://s1341.photobucket.com/user/nebula-314/media/20130520_221430_zps840432f0.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20130520_221430_zps840432f0.jpg[/URL][/PLAIN]

[URL=http://s1341.photobucket.com/user/nebula-314/media/20130520_222748_zps3f68f545.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20130520_222748_zps3f68f545.jpg[/URL][/PLAIN]

This is how i set it up. If this set up is correct I am having trouble evluating the tan(25+theta) part. If there is an easier way please help.

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sorry for the verticle photo.

SammyS
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I am curious if I am setting this problem up right.

Three forces act on the bracket. Determine the magnitdue and direction theta of F2 so that the resultant force is directed along the positive u axis and has a magnitude of 50lb.

[ IMG][ URL=http://s1341.photobucket.com/user/nebula-314/media/20130520_221430_zps840432f0.jpg.html][ [Broken] IMG]http://i1341.photobucket.com/albums/o745/nebula-314/20130520_221430_zps840432f0.jpg[/URL] [Broken]

[ IMG][ URL=http://s1341.photobucket.com/user/nebula-314/media/20130520_222748_zps3f68f545.jpg.html][ [Broken] IMG]http://i1341.photobucket.com/albums/o745/nebula-314/20130520_222748_zps3f68f545.jpg[/URL] [Broken]

This is how i set it up. If this set up is correct I am having trouble evluating the tan(25+theta) part. If there is an easier way please help.
You could use the angle sum identity for tangent: ##\displaystyle \ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\ .##

But I think it would be better to solve for tan(25° + θ), then take the arctan of both sides to get a numerical result for 25° + θ . (Make sure that's in degrees.) Subtract 25° from that.

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ahhhhh ok thanks, ill try this in the morning.