direction will be found using
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depending on the quadrant involved.
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.
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:
V_{1x} = 5cos(90°) = 5*0 = 0 (it's going due north, not east or west even a little)
V_{1y} = 5sin(90°) = 5*1 = 5 (it would be -5 if we were headed south)
now we figure out how much the wind is contributing:
V_{2x} = 6cos(45°) = 6*(0.707) = 4.242
V_{2y} = 6sin(45°) = 6*(0.707) = 4.242
we add the V_{x}'s and the V_{y}'s separately:
V_{1x} + V_{2x} = 4.242 <--total of components in the x-direction (east)
V_{2x} + V_{2y} = 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).