# Math Help - Forces as a Vectors

1. ## Forces as a Vectors

Find the horizontal and vertical components of each of the forces:

f) 36N acting vertically

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

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

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

d) forces of 7N east and 12N north

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

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

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

My Answer: 10N, N8°W --------- is my answer correct?

g) forces of 6N southeast and 8N northwest

My Answer: 2N, N45°W ---------------- is my answer correct?

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

f) 36N acting vertically

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

How do I find the $y$ value for the component if I cannot solve for $36sin \theta$?
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, $\theta = 90^o$.)

-Dan

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

d) forces of 7N east and 12N north

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

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

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.

Add them by components:
Using the usual xy coordinate plane, the first is (7, 0) and the second is (0, 12). So the resultant vector will be (7 + 0, 0 + 12) = (7, 12).

So the magnitude is $r = \sqrt{7^2 + 12^2} = 13.892443989$ 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.)

-Dan