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September 7th, 2008, 12:40 PM
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The problem goes as follows :
Two forces Fsub1 and Fsub2 with magnitudes 10lb and 12lb act on an object at a point P as shown in the figure. Find the resultant force F acting at P as well as its magnitude and it's direction. (Indicate the direction by finding the angle theta shown in the figure).
How would I even start this? I'm sure someone can just give me the answer but i'd like to know how to get there. Thanks in advance. |
September 7th, 2008, 12:44 PM
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Resolve the forces into their horizontal and vertical components (check the parallelogram rule and apply that to a rectangle whose sides are horizontal and vertical).
Then add the horizontal forces together (one will be opposite the other so you'd add a negative force to a positive) and add the vertical forces together to get the horiz and vert components of the resultant force. Then add those together in the vector style and you're done.
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September 7th, 2008, 01:10 PM
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 HUH? Sorry I just looked at what you wrote for a while and don't understand it. |
September 7th, 2008, 01:35 PM
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How do I get the horizontal and the vertical components? |
September 7th, 2008, 01:41 PM
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Quote:
Originally Posted by JonathanEyoon  How do I get the horizontal and the vertical components? | Trigonometry.
Fsub2 horiz component is fsub2 cos 30, vert component fsub2 sin 30, for example.
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September 7th, 2008, 02:05 PM
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Quote:
Originally Posted by Matt Westwood Trigonometry.
Fsub2 horiz component is fsub2 cos 30, vert component fsub2 sin 30, for example. |
Ah I think I see now. So to get the horizontal component of Fsub2, you did Cos since cos = adjacent / hypotenuse? Similarly you did the same for the vertical which is opposite / hypotenuse? Ok so by this, I got
Fsub1 = -10cos45i + 10sin45j
Fsub2 = 12cos30i + 12sin30j
Is this right? If so what should I do from here? |
September 7th, 2008, 03:12 PM
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Another question is, What do I do with the lbs they have listed in the problem? |
September 7th, 2008, 03:14 PM
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Quote:
Originally Posted by JonathanEyoon Ah I think I see now. So to get the horizontal component of Fsub2, you did Cos since cos = adjacent / hypotenuse? Similarly you did the same for the vertical which is opposite / hypotenuse? Ok so by this, I got
Fsub1 = -10cos45i + 10sin45j
Fsub2 = 12cos30i + 12sin30j
Is this right? If so what should I do from here? |
Add them up (converting your sines and cosines into numbers first).
Then do the reverse of what you just done getting the magnitude by pythagoras and the angle by arctan.
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September 7th, 2008, 03:15 PM
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Quote:
Originally Posted by JonathanEyoon Another question is, What do I do with the lbs they have listed in the problem? | The forces is in lbs (prehistoric units, damn them to hell) so the resultant force is also in lbs.
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September 7th, 2008, 03:42 PM
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