I have a question which requires finding the magnitude and direction of a resultant force (including angles as to be able to accurately draw on paper). The question is written as follows:
"consider two forces, F1 and F2, of magnitudes 17N and 9N respectively, acting on a particle (diagram). what is the magnitude and direction of the resultant forces?"

I have attached the drawing given.

Any help would be appreciated, preferably with working out so I can see how it is done for any future reference.

Cheers, -Paaul

Here's the process:

First break each of the forces (F1 and F2) into their horizontal and vertical components (x-direction and y-direction):

F1x = F1 *cos(27)
F1y = F1* sin(27)
F2x = -F2*cos(54)
F2y = F2*sin(54)

Note the negative sign - this is because the x-component of F2 is in the negative x direction.

Now add the x-components and y-components:

Fx_total = F1x + F2x
Fy_total = F1y + F2y

Then you combine these two force vectors into a single resultant force vector, F_total:

Magnitude of F_total = sqrt(Fx_total^2 + Fy_total^2)

Angle of F_total = Atan(Fy_total/Fx_total). Pay attention to the sign here, and make sure you pick the correct quadrant for the final answer.

Post back with hat you get for a final answer.

Quote:

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Originally Posted by ebaines

Here's the process:

First break each of the forces (F1 and F2) into their horizontal and vertical components (x-direction and y-direction):

F1x = F1 *cos(27)
F1y = F1* sin(27)
F2x = -F2*cos(54)
F2y = F2*sin(54)

Note the negative sign - this is because the x-component of F2 is in the negative x direction.

Now add the x-components and y-components:

Fx_total = F1x + F2x
Fy_total = F1y + F2y

Then you combine these two force vectors into a single resultant force vector, F_total:

Magnitude of F_total = sqrt(Fx_total^2 + Fy_total^2)

Angle of F_total = Atan(Fy_total/Fx_total). Pay attention to the sign here, and make sure you pick the correct quadrant for the final answer.

Post back with hat you get for a final answer.

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Ok,

for F1x I got = -4.97
for F1y I got = 16.26
for F2x I got = 7.46
for F2y I got = -5.03

Fx total = 2.49
Fy total = 11.26

magnitude = 11.53

and angle I get to Atan 4.52
what does the A respresent before tan?


Thanks.

First mistake - you seem to have used radians rather than degrees when you determined the values of the sines and cosines. If you look at the diagram it should be obvious that the horizontal component of F1 is positive, since it points to the right - which should tell you right away that you have an error. Same thing with F2y - clearly it should be a positive value.

"Atan" means Arctan.

Thanks again for the help, I'll give more feedback tomorrow, I'm using Google calculator to get the answers as my scientific calculator is at home (obviously a bad decision lol).

-Paaul

Quote:

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Originally Posted by ebaines

First mistake - you seem to have used radians rather than degrees when you determined the values of the sines and cosines. If you look at the diagram it should be obvious that the horizontal component of F1 is positive, since it points to the right - which should tell you right away that you have an error. Same thing with F2y - clearly it should be a positive value.

"Atan" means Arctan.

---

Ok, I've now got a proper calculator and its looking better.

for F1x I got = 15.15 (2dp)
for F1y I got = 17.72 (2dp)
for F2x I got = -5.29 (2dp)
for F2y I got = 7.28 (2dp)

Fx total = 9.86
Fy total = 15

magnitude = 17.95

angle = 56.66 (2dp)

This looks more along the correct lines.

-paaul

Nice one, Cheers for the help.

I have other examples I will be needing help on, feel free to help when I post them :)