Setting up Forces: Finding Theta of F2

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The discussion focuses on determining the angle theta of force F2 to ensure the resultant force on a bracket is directed along the positive u axis with a magnitude of 50 lb. The user expresses uncertainty about their setup and specifically struggles with evaluating the tangent of the angle sum, tan(25° + theta). A suggested approach involves using the angle sum identity for tangent to simplify the calculation. It is recommended to solve for tan(25° + theta) first and then apply the arctangent function to find the angle. The user plans to try this method the following day.
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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.

[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.
 
whynot314 said:
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][ IMG]http://i1341.photobucket.com/albums/o745/nebula-314/20130520_221430_zps840432f0.jpg[/URL]


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

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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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