# Forces as a Vectors

• Feb 14th 2008, 11:41 AM
Macleef
Forces as a Vectors
Find the horizontal and vertical components of each of the forces:

f) 36N acting vertically

$\displaystyle {\overrightarrow {r} } = (0, 36sin \theta)$

How do I find the $\displaystyle y$ value for the component if I cannot solve for $\displaystyle 36sin \theta$?

Find the resultant of each pair of forces acting on an object.

d) forces of 7N east and 12N north

$\displaystyle \sqrt {|{\overrightarrow {r} }|^2}$ = $\displaystyle \sqrt{7^2 + 12^2}$
$\displaystyle |{\overrightarrow {r} }| = 13.9$

$\displaystyle tan \theta = \frac {12}{7}$
$\displaystyle \theta = 60$°

Therefore, 13.9N, _____________ (How do I find the direction of this force? Can you please show me how using a diagram?)

e) forces of 6N southwest and 8N northwest

g) forces of 6N southeast and 8N northwest

• Feb 14th 2008, 03:09 PM
topsquark
Quote:

Originally Posted by Macleef
Find the horizontal and vertical components of each of the forces:

f) 36N acting vertically

$\displaystyle {\overrightarrow {r} } = (0, 36sin \theta)$

How do I find the $\displaystyle y$ value for the component if I cannot solve for $\displaystyle 36sin \theta$?

You are over-thinking it. The vector is vertical, hence it is acting entirely in the y direction. So the vector is (0, 36). (If you want, $\displaystyle \theta = 90^o$.)

-Dan
• Feb 14th 2008, 03:17 PM
topsquark
Quote:

Originally Posted by Macleef
Find the resultant of each pair of forces acting on an object.

d) forces of 7N east and 12N north

$\displaystyle \sqrt {|{\overrightarrow {r} }|^2}$ = $\displaystyle \sqrt{7^2 + 12^2}$
$\displaystyle |{\overrightarrow {r} }| = 13.9$

$\displaystyle tan \theta = \frac {12}{7}$
$\displaystyle \theta = 60$°

Therefore, 13.9N, _____________ (How do I find the direction of this force? Can you please show me how using a diagram?)

You pretty much have it.

So the magnitude is $\displaystyle r = \sqrt{7^2 + 12^2} = 13.892443989$ and the angle (in the first quadrant, since the x and y components are both positive) will be 59.743562836 degrees. (Or 60. degrees N of E.)