Find the magnitude and direction of the equilibrant (resulting force) of each of the following forces: 16 N and 10 N acting at an angle of 10 degrees to each other.

Please make it as visual as possible... Thanks! (best answerer gets a thanks ^^)

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- Apr 18th 2009, 06:35 PMx y z[SOLVED] Easy Vectors Question
Find the magnitude and direction of the equilibrant (resulting force) of each of the following forces: 16 N and 10 N acting at an angle of 10 degrees to each other.

Please make it as visual as possible... Thanks! (best answerer gets a thanks ^^) - Apr 18th 2009, 11:07 PMearboth
There are different ways to solve this question:

1. Draw the parallelogram of forces. Use the Cosine rule to calculate the length of the larger diagonal. Keep in mind that the adjacent sides of the parallelogram include an angle of 180° - 10° = 170° :

$\displaystyle F_r = \sqrt{10^2+16^2-2 \cdot 10 \cdot 16 \cdot \cos(170^\circ)}$

2. Use vectors:

Let $\displaystyle \overrightarrow{F_1} = (10,0)$ and

$\displaystyle \overrightarrow{F_2} = 16(\cos(10^\circ), \sin(10^\circ))$

Then

$\displaystyle \overrightarrow{F_r} = (10,0) + 16(\cos(10^\circ), \sin(10^\circ)) = (10+16 \cos(10^\circ), 16 \sin(10^\circ))$

Now calculate the magnitude of $\displaystyle \overrightarrow{F_r}$ :

$\displaystyle | \overrightarrow{F_r} | = \sqrt{\left(10+16 \cos(10^\circ) \right)^2 + \left(16 \sin(10^\circ) \right)^2}$ - Apr 19th 2009, 05:29 AMx y z
Thanks.

But how do I then get the direction of the vector? - Apr 19th 2009, 11:47 AMearboth
If you have the vectors $\displaystyle \vec u$ and $\displaystyle \vec v$ then the included angle is calculated by:

$\displaystyle \cos(\angle(\vec u , \vec v)) = \dfrac{\vec u \cdot \vec v}{|\vec u| \cdot |\vec v|}$

With your vectors you get:

$\displaystyle

\cos(\angle(\overrightarrow{F_r} , \overrightarrow{F_1}) = \dfrac{(10+16 \cos(10^\circ), 16 \sin(10^\circ))

\cdot (10, 0)}{2\cdot \sqrt{80 \cdot \cos(10^\circ)} \cdot 10}$

After your calculator has cooled down a little bit you'll get:

$\displaystyle \angle(\overrightarrow{F_r} , \overrightarrow{F_1}) \approx 6.157^\circ$

Keep in mind that you now have calculated the angle which is included by $\displaystyle \overrightarrow{F_r} \text{ and } \overrightarrow{F_1}$ - Apr 19th 2009, 02:56 PMx y z
That's what I got too, but the answer says that it's 174 degrees to the 10N vector. Why did they subtract from 180 degrees?

- Apr 19th 2009, 10:18 PMearboth
- Apr 20th 2009, 11:42 AMx y z
Wow, thank you so much! <3

- Apr 21st 2009, 09:59 PMearboth
I finally found out where the answer in your textbook comes from:

The most important word in your question is**equilibrant (resulting force)**.

If you combine the given forces to a resultant force and you should find a third force such that the complete system is in equilibrium then the result must be a force with the magnitude of $\displaystyle \overrightarrow{F_1}+\overrightarrow{F_2}$ and the opposite direction. That is $\displaystyle -\left( \overrightarrow{F_1}+\overrightarrow{F_2} \right)$.

I've attached a sketch. You easily find the angle in question.