# [SOLVED] Easy Vectors Question

• Apr 18th 2009, 07:35 PM
x 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 19th 2009, 12:07 AM
earboth
Quote:

Originally Posted by x y z
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.

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° :

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

2. Use vectors:

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

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

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

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

$| \overrightarrow{F_r} | = \sqrt{\left(10+16 \cos(10^\circ) \right)^2 + \left(16 \sin(10^\circ) \right)^2}$
• Apr 19th 2009, 06:29 AM
x y z
Thanks.
But how do I then get the direction of the vector?
• Apr 19th 2009, 12:47 PM
earboth
Quote:

Originally Posted by x y z
Thanks.
But how do I then get the direction of the vector?

If you have the vectors $\vec u$ and $\vec v$ then the included angle is calculated by:

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

$
\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:

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

Keep in mind that you now have calculated the angle which is included by $\overrightarrow{F_r} \text{ and } \overrightarrow{F_1}$
• Apr 19th 2009, 03:56 PM
x 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, 11:18 PM
earboth
Quote:

Originally Posted by x y z
...
Please make it as visual as possible... ...

Quote:

Originally Posted by x 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?

I've attached the sketch of the situation ... and honestly I can't explain where the result in your textbook comes from, sorry!
• Apr 20th 2009, 12:42 PM
x y z
Wow, thank you so much! <3
• Apr 21st 2009, 10:59 PM
earboth
Quote:

Originally Posted by x 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?

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 $\overrightarrow{F_1}+\overrightarrow{F_2}$ and the opposite direction. That is $-\left( \overrightarrow{F_1}+\overrightarrow{F_2} \right)$.

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