How Do You Calculate Resultant Forces Using Geometric Vectors?

[Macleef]

[SOLVED] Geometric Vectors

Homework Statement

Find the magnitude and the direction (to the nearest degree) of the resultant of each of the following systems of forces.

a. forces of 3 N and 8 N acting at an angle of 60º to each other - use geometric vectors to solve this problem

Answer: 9.8 N, 15º to 8N

Homework Equations

The cosine and sine laws.

r = a^2 + b^2 - 2(a)(b)cosr

sinθ/a = sinθ/b = sinθ/r

The Attempt at a Solution

I tried using the sine law first, but I can't retrieve the angle to find the right resultant using the cosine law.

 

[awvvu]

I'd break the 3N and 8N forces into their x and y components and add them, unless that doesn't count as "geometric vectors"

 

[Macleef]

I'd break the 3N and 8N forces into their x and y components and add them, unless that doesn't count as "geometric vectors"

You're describing algebraic vectors. . .Even if I do it this way, I don't know how to get the angle to substitute in the formula:

(a, b) = (#cosθ, #sinθ)

 

[awvvu]

Well, choose an easy angle for one of them, like 0 deg for the 3N force, then the 8N force must be at 60 deg.

Or, if you want to find theta in your diagram, just use law of sines.

$$\frac{\sin(\theta)}{8} = \frac{\sin(60 \deg)}{3}$$

 

[Macleef]

What? I don't get it what you mean?

 

[awvvu]

Unless I'm misunderstanding something, if the two vectors are 60 deg from each other, one possible angle for them are 0 and 60 degrees. Or 10 and 70 degrees. Or whatever you want.

There must be something more to this problem because the resulting vector could be in any direction, not just 15 deg.

 

[Macleef]

I tried doing sine law, but I keep getting error.

And the angles you've provided don't give the right resultant.

The only angles that do give me approximately 9.8 N are 118, 119 and 120 degrees.

But, I want to know how to solve this problem without looking at the answers because my test is coming up soon and I don't have anything to refer to. . .

 

[awvvu]

Oh, that's because 8/3 sin(60deg) = 2.3... which shouldn't happen. There's something wrong with your diagram.

The 60 degree angle is opposite the 3 side, while theta is opposite 8, which isn't possible.

There's something more to this problem because saying the two vectors are 60 degrees apart is not enough to get that answer.

 

[Macleef]

Then how would you graph the problem? I don't understand the wording. . .

 

[HallsofIvy]

You say the problem is: "a. forces of 3 N and 8 N acting at an angle of 60º to each other - use geometric vectors to solve this problem"

But in your figure you have the 8 N force 60 degrees from the resultant.

If the two forces are 60 degrees apart, then the top of your triangle (where "head" of 8 N force meets "tail" of 3 N force) is 180- 60= 120 degrees.