# Can you find the resultant force between these two vectors?

Determine the magnitude of the resultant force on an object if force $A$ is pulling the object with $150$ lbs of force and force $B$ is pulling with $300$ lbs, and the angle between the two forces is $110^\circ$.

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would you be willing to tell me the steps you take to find the resultant force? – Elle May 17 '13 at 10:25
I am not worried at all that mathematicians can answer this, but I feel like this belongs to physics.stack.exchange. Moderators? – Patrick Da Silva May 17 '13 at 10:54
Actually im studying for a trig final in college. I keep doing something wrong. i get 252.6 for the magnitude that iostream posted and I'm getting 285.9 from my formula, which is r^2 = a^2 + b^2 - 2(a)(b)cos(180 - θ) – Elle May 17 '13 at 10:58
@iostream007, I think your equation should be $|150+ 300\times \cos 110^\circ|$ because the negative will come from $\cos$. – Tpofofn May 17 '13 at 11:01

1) Draw a diagram with coordinate axes:

2) Determine the components of the two forces:

$${F_A}_x = F_A = 150 lbs \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space {F_A}_y = 0$$ $$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space{F_B}_x = -F_B\cos70° = -103 lbs \space\space\space\space\space\space\space\space\space\space{F_B}_y = F_B\sin70° = 282 lbs$$ 3) Find the components of the resultant force $$R_x = {F_A}_x + {F_B}_x = 47 lbs$$ $$R_y = {F_A}_y + {F_B}_y = 282lbs$$ 4) Apply Pythagoras' theorem to get the magnitude of the resultant: $$R = \sqrt {R_x^2 + R_y^2} = 286lbs$$

If you want to know more, here's a good article which explains each step in more detail: How to find the resultant force acting on an object

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use this formula of triangle $a^2=b^2+c^2-2bc\cdot \cos A$ where a,b,c are sides of triangle.

so resultant force $$\vec R^2={150}^2+{300}^2-2\cdot150\cdot 300 \cos 70^\circ$$ $$\vec R=\sqrt {22500+90000-30781.81}$$ $$\vec R=285.86\,lbs$$

There was typo in the formula which I have now corrected. a^2=b^2+c^2 - 2bc * cos A

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If we assume $f_A$ points in the positive $x$ direction and $f_B$ is located in quadrant II, we have,

$$f_A = \left[ \begin{array}{c} 150\\ 0 \end{array} \right]$$

and

$$f_B = \left[ \begin{array}{c} 300\cos(110) \\ 300\sin(110) \end{array} \right]$$

The resultant is

$$f_R = f_A + f_B = \left[ \begin{array}{c} 150 + 300\cos(110) \\ 300\sin(110) \end{array} \right]$$

then

$$\|f_R\| = \sqrt{(150 + 300\cos(110))^2 + (300\sin(110))^2}=285.86$$

Note: Notice that $\cos(110)$ is negative.

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mathwarehouse.com/vectors/resultant-vector.php check this. your angle $110^\circ$ is wrong because it is angle between vectors but when we use triangle law to find resultant we place vectors in such a way that first vectors head and second vectors tail should be on same point so correct angle will be $70^\circ$ – iostream007 May 20 '13 at 8:47
@iostream007, I did not use the triangle law. I realized the vectors in a coordinate system and then added them by component. – Tpofofn May 21 '13 at 10:43
but your answer is not correct – iostream007 May 22 '13 at 18:22
@iostream007, I get the same answer as your approach with the law of cosines. Just a different method. – Tpofofn May 23 '13 at 3:10

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