1. Vector Problem

Two forces with magnitudes of 200 and 100 pounds act on an object at angles of 60° and 170° respectively. Find the direction and magnitude of these forces. Round to two decimal places in all intermediate steps and in your final answer.

Can anyone help me solve this? Thank you for any help you can provide.

2. Re: Vector Problem

$R_x = F_{1x} + F_{2x}$

$R_y = F_{1y} + F_{2y}$

$|R| = \sqrt{R_x^2 + R_y^2}$

direction will be found using

$\arctan\left(\frac{R_y}{R_x}\right)$ ,

3. Re: Vector Problem

So I would have Rx=200 + 60
Ry=100 + 170

Is that right?

4. Re: Vector Problem

why are you adding the force magnitudes with their angle measures? have you dealt w/ adding/subtracting vectors at all?

5. Re: Vector Problem

Not really no. This is my first lesson with all this and I'm like completely confused. I must have missed that formula when I was taking my notes down. I have like 3 pages of notes and can't find that formula anywhere.

7. Re: Vector Problem

Thanks for the video. But I'm still really confused. What is F in the formula?

8. Re: Vector Problem

Originally Posted by Itel
Thanks for the video. But I'm still really confused. What is F in the formula?
Two forces with magnitudes of 200 and 100 pounds act on an object at angles of 60° and 170° respectively.
$F_1$ and $F_2$ represent the two forces stated in your problem ... since they have magnitude and direction, both are vectors.

9. Re: Vector Problem

Ok, so F1x+F2x=100+200=300

So Rx=200

What is Fy?

10. Re: Vector Problem

Originally Posted by Itel
Ok, so F1x+F2x=100+200=300
no.

$F_{1x} = 200\cos(60^\circ)$

$F_{2x} = 100\cos(170^\circ)$

$R_x = 200\cos(60^\circ)+100\cos(170^\circ)$

now do the y-components using the correct trig ratio ...

11. Re: Vector Problem

if you have a vector given in terms of "B amount of force in direction angle A", the force is the "x-direction" is Bcos(A), and the force in the "y-direction" is Bsin(A).

here, "x-direction" is "the reference direction you are measuring your angle A from".

once you have "x-direction" and "y-direction" components, to sum the forces, you sum each component, separately (don't add "x-direction" stuff to "y-direction" stuff, it's "bad math").

after you're done with "adding of the forces", you use the formulas skeeter gave in his post, to convert back to "direction and angle".

here is a simple example:

a sailboat is moving due north at 5 knots/hour. the wind is blowing northeast at 6 knots/hour. what is the resulting direction and speed of the boat (assuming no course correction)?

let's pick "due east" as our "reference angle 0". so east/west is the x-axis, north/south is the y-axis.

first, we're going to "resolve" the velocity vectors into x and y components. for the boat, this is easy:

V1x = 5cos(90°) = 5*0 = 0 (it's going due north, not east or west even a little)
V1y = 5sin(90°) = 5*1 = 5 (it would be -5 if we were headed south)

now we figure out how much the wind is contributing:

V2x = 6cos(45°) = 6*(0.707) = 4.242
V2y = 6sin(45°) = 6*(0.707) = 4.242

we add the Vx's and the Vy's separately:

V1x + V2x = 4.242 <--total of components in the x-direction (east)
V2x + V2y = 9.242 <--total of components in the y-direction (north)

now, we have to figure out the magnitude of the resultant forces.

first we square the total components, and add the result:

(4.242)2 + (9.242)2 = 17.995 + 85.415 = 103.41

then we take the square root:

√103.41 = 10.17 knots/hour <---this is the resulting speed (the wind really picked things up!)

now to find the direction we're headed:

arctan(9.242/4.242) = arctan(2.179) = 65.35 degrees (this is sort of north-northeast), which is to say 24.65 degrees off-course!

(you might want to check my arithmetic, i make mistakes like ordinary humans).