What is the magnitude of the resultant force and angle in this vector problem?

[recoil33]

Okay, I don't really understand vectors exactly - I think i do, but then i confuse myself.

"Two forces have magnitudes of 26 N and 19 N and the angle between them is 125°. Find the magnitude of the resultant and the angle it makes with the larger of the two forces. Give answers correct to 1 decimal place."

Q1. Resultant Force(N):

Q2. Angle(theta):



Alright, i used the "Parallelogram Law"

Then ended up solving for:

|R²| = 26² + 19² - 2*(26)*(19)*Cos(125)

|R| = (26² + 19² - 2*(26)*(19)*Cos(125))^(1/2)

|R| = 40.046
|R| = 40.0N

Although, my answer is wrong. Apparently I've used the wrong angle, something to do with a angle complementary with 180 degrees?

Any help will be appreciated.

(Can't solve Q2 without the answer to Q1) - I should be alright there.

 

[Fewmet]

Okay, I don't really understand vectors exactly - I think i do, but then i confuse myself...

Although, my answer is wrong. Apparently I've used the wrong angle, something to do with a angle complementary with 180 degrees?

Any help will be appreciated.

(Can't solve Q2 without the answer to Q1) - I should be alright there.

Your problem is more with trig than vectors. Did you sketch the situation? In the Law of Cosines, angle C has to be opposite the longest side (the diagonal that cuts the 125º in this case).

Use the sketch to see what angle is opposite the diagonal and swap it for 125º in your calculation.

 

[Omega_Prime]

Although, my answer is wrong

Hmm, I just gave this a shot myself and got the same magnitude for the r vector, it looks correct to me...!

What did you get for the angle?

 

[mege]

I get something different.

Converting the forces to vectors and then adding them, we can attempt verify the result. Let...
A = Origin
B = 19N 'point' along the x-axis
C = 26N 'point' in the 2nd quadrant 125deg from the x-axis.
D = Resultant 'point'
We can find their coordinates by using unit-circle equivalencies:

AB = <19cos 0, 19sin 0> = <19,0>
AC = <26cos 125, 26sin 125> ~= <-14.913,21.298>

AB + AC = AD ~= <4.087,21.298> |AD| ~= 21.6866

Angle between AD and the x-axis is arctan 21.298/4.087 = 79.1372deg

So, the final answer = 21.7 N at an angle of 79.1deg from the smaller force-vector.

Is 125deg the proper angle to put into the parallelogram law? I think that is where your approach is not correct. 'Tip to tail' graphing will help find the proper angle using the parallelogram law.

 

[recoil33]

Thanks everyone,

I realized i was using the wrong angle, i think i confused myself with putting the force of 26N 125 degrees away in the other direction. Which should not matter, although it confused myself.

Anyways, I solved it finally.

As for Q2.

I used the sin rule,
a/SinA = b/SinB

19/sin(theta) = |R|/Sin55

= (19Sin55)/21.687 = Sin(x)
= 45.9 Degrees

 

[mege]

Thanks everyone,

I realized i was using the wrong angle, i think i confused myself with putting the force of 26N 125 degrees away in the other direction. Which should not matter, although it confused myself.

Anyways, I solved it finally.

As for Q2.

I used the sin rule,
a/SinA = b/SinB

19/sin(theta) = |R|/Sin55

= (19Sin55)/21.687 = Sin(x)
= 45.9 Degrees

Yep - I just used the x-axis/smaller force for my angle. You'll notice that:
125deg - your answer = my answer (and verifys the correctness)