Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $110^\circ$.

share|cite|improve this question
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 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:

free body diagram with coordinate axes

2) Determine the components of the two forces:

x y 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

share|cite|improve this answer

enter image description here

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

share|cite|improve this answer
downvoters please leave a comment – iostream007 May 20 '13 at 8:47

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

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


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

The resultant is

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


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

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

share|cite|improve this answer 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

protected by T. Bongers Apr 10 '14 at 0:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.