Forces with magnitudes of 125 N and 300 N are acting on a hook...(See below for the rest of the question...)?

Question

The angle between the two forces is 45 degrees... I'm to find the magnitude AND the direction of the resultant of those two forces. I have the answer, but I need to see how it's found.

I know this is a simpler problem because I took a year of college physics because I'm a premed, but it was a while ago and I've moved on in and almost forgotten it. I have some major brushing up to do before grad exams, but can anyone help me solve this?

Answer 1

take a ruler and a pencil, and draw the two forces as two lines originated in the same point having the forces' magnitude length and an angle of 45 between the lines. put a little arrow at the end of each line.
Now you should make a parallelogram using the two forces, as you move one of the force along the other one till at the end point of it. The opposite sides of a parallelogram are parallel to each other and the same length, so take care how you draw. Now unite the origin of the two forces with the opposite vertex of the parallelogram, and point it with an arrow to this opposite vertex. This will be your resultant force, and it is it's direction.
For the magnitude you should use Pythagoras's law.
C^2 = A^2 + B^2 - 2AB cos alfa
alfa is the angle opposite to the side C, between A and B.
Let you mark the two forces as A and B, C is the resultant.
You can see in your parallelogram that there is a triangle containing all these lines. The angle between the two sides of the triangle opposite to the C side is 135.
cos 135 = - sin 45
C^2 = 125^2 + 300^2 +2 x 125 x 300 x sin 45 =
= 15625 + 90000 + 75000 x 0.7 = 158658.0086
C = 398.31 N

Answer 2

398 N @ 32 degrees

Answer 3

Draw two forces at any angle from horizontal reference but 45 degrees apart. Let the right force be 125N and the left be 300N forces. If you use the graphical or parallelogram method and use some gemetric manipulations, you can find that the resultant force will be 398.32N at 12.82 degrees from the 300N force or at 32.18 degrees from the 125N force. (what a pitty I could not draw here).

Answer 4

It's been almost 20 years since I've been in school and I'm sure my method isn't the most elegant but here's what I came up with. I don't know how much you know so bear with me.
You're adding vectors here in this case force vectors. The standard way to represent vectors is with a line segment with an arrow at one end such that the length of the segment is the magnitude of the vector and the angle of the segment on a graph represents the direction. the head of the vector is the end with the arrow.
What I did is to place the 300n vector horizontally on a standard 2 dimension x/y graph on the x axis with the tail at the origin. Then to add the two vectors together place the tail of the second 125n vector at the head of the first at the correct angle in your case a 45 degree positive slope up. The sum of these two is a vector that starts at the tail of first vector and finishes at the head of the second vector.
From here I'm sure there is an easier way to find the sum, but here's what I did.
I made a right triangle by dropping a segment vertically down from the head of the second vector to the x axis.
I took the sin of 45 degrees which is going to be the ratio of the magnitude of the second vector (125) to that line segment I just drew. .707
multiplied .707 and 125 to get the length of the segment 88.38.
Since this little triangle is a 45 -45 right triangle I know the other segment is the same length.
From here I can i can figure out the magnitude of the summed vector with good ol Pythagoras
The square root of ((300+88.38)^2 + 88.38^2) which is 398.3 approximately
lastly I took the inverse tangent of 88.38/388.38 to get the angle
or tan^-1 of .227 or 12.8 degrees
Then just use your graph to see that the angle is 12.8 degrees relative to the larger vector