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coolkat4
Two forces with magnitudes of 150 and 75 pounds act on an object at angles of 30° and 150° respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.
it looks like you asked this 3 days ago and didnt get a response. Im not sure what stamp is refering to since I see no posts in either of your profiles that would suggest such a response :/
i personally would use the law of cosines for the magnitude \[c^2=a^2+b^2-2ab~cosC\] \[c=\sqrt{a^2+b^2-2ab~cosC}\]
the direction can be obtained in a few ways. prolly converting the angles to rect coords is the most outright
See image. You can also try to get the x- and y-components of both forces (red vectors). There are two 30-60-90 triangles in this situation, so the components are readily known. (black horizontal and vertical vectors). Only after this, when you have to find the magnitude and angle of the resultant force (blue vector), you have to use the Pythagorean Theorem and a sin, cos or tan, and this will make rounding off necessary.
i just needed guidance in the right direction not the answer
\[|c|=\sqrt{150^2+75^2-2(150)(75)~cos(150-30)}\] http://www.wolframalpha.com/input/?i=sqrt%28150%5E2%2B75%5E2-2%28150%29%2875%29cos%28150-30%29%29
I think you are both wrong... @coolkat4: You don't give a calculation so I don't know what you did. @amistre64: Using the law of cosines in this way gives the length of the segment between the tips of both vectors (=AB in my drawing). I used the Pythagorean Theorem on the x- and y- resultants: \[|F_r|=\sqrt{(112.5)^2+ (37.5\sqrt{3})^2}=\sqrt{16875}=75\sqrt{3} N\]
.... yeah, I was reading it incorrectly :) thanx
by the way, if we turn one of the vectors 180 we can use the law of cosines ;) |dw:1361979481180:dw| \[\sqrt{150^2+75^2-2(150)(75)~cos60^o}=75\sqrt{3}\]