# Two force systems have the same resultant

1. Mar 12, 2014

### bawse

1. The problem statement, all variables and given/known data

Attached image.

2. Relevant equations

Cosine rule
|F| = √(Fx2 + Fy2)

3. The attempt at a solution

I tried to calculate the force components in the x and y direction for both and got two expressions for R (one from each system) and made them equal each other. However I have two unknowns (F1 and F2) when I try this and am unsure how to proceed :/

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2. Mar 12, 2014

### Simon Bridge

You need two equations - the second equation comes from finding the angle.

3. Mar 12, 2014

### bawse

Im assuming by angle you mean the angle at which the resultant acts to the horizontal? I understand that but I'm having difficulty with calculating the resultant to begin with. Was my method to equate the two resultant equations from each system the correct method?

4. Mar 12, 2014

### Simon Bridge

Yeah - the magnitudes and the directions have to be the same - so you'll have one relation equating the magnitudes and a second one equating the angles - probably the sine cosine or tangent of the angles.

Start by writing the two relations down next to each other.

5. Mar 12, 2014

### bawse

I get the following relation for magnitude:

F12 + 46.42F1 + 3025 = 900 + F22 + 20.5F2

For the angle relations, I got confused as to what sides to use in the sine rule but here is what I did:

First force system:

R/sin115 = 55/sin β

Second force system:

R/sin110 = F2/sin β

Please excuse if I'm doing it wrong, its just that my lecturer has to go through material very quickly as I'm in an accelerated class and so has minimal time to explain everything.

Thanks

6. Mar 12, 2014

### Simon Bridge

OK I see where you are coming from.

Looking at how you are thinking, you may be best to just find F1 and F2 by equating the components between the two triangles. I suspect you tried that already - show me.

This should give you two equations with two unknowns.
Write those next to each other :)

From there you should be able to find everything else.

Note: if I put u as the 55N force and v as the 30N force, I'm drawing the triangles as:

$\vec R = \vec F_1+\vec u = \vec v + \vec F_2$

7. Mar 12, 2014

### bawse

What do you mean by the components between two triangles? Do I equate the x and y components from each triangle (i.e Fx=Fx) together or the sum of the x and y components from each triangle which is what i did in my previous post? ( i.e Fx^2 +Fy^2 = Fx^2 + Fy^2)

EDIT: I posted this on one of the student forums we have and got this response:

"Simultaneous Equations.

Because the resultants of both individual systems are identical, the total x and y components of each individual system are also equal. Using the X and Y forces you can form two equations with (some constant)F1 +/- z(some number) = F2 or vice versa. Then by substituting either F1 or F2 in terms of the second variable from one equation into your other equation you can solve for one unknown. Then go back to one of your earlier equations and solve for the other unknown. From there its pretty straightforward. Hope this helps!"

Is this along the same lines as what you're trying to say? Thanks for the help, i'll be sure to finish this question off once I get home again.

Last edited: Mar 12, 2014
8. Mar 12, 2014

### Simon Bridge

Yep. You know the resultant is the same for both sets of forces - that means that the x-component of the resultants are the same and the y components of the resultants are the same.

i.e.
$R_x = F_{1x}+u_x = F_{2x}+v_x\\ R_y = F_{1y}+u_y = F_{2y}+v_y$

... it turns out you know all the angles you need to do that.

9. Mar 12, 2014

### bawse

I tried doing this just now and I get F1x in terms of F2x. What do I do from here? Which equation do I substitute this into?

Last edited: Mar 12, 2014
10. Mar 12, 2014

### Simon Bridge

Don't forget to use trigonometry to put $F_{ix}$ and $F_{iy}$ in terms of $F_i$ ($i\in\{1,2\}$).

11. Mar 12, 2014

### bawse

I think I finally solved it :D

My answer is a bit off, probably due to rounding but I have attached my working, could you please check and see if there was a faster way to solve the problem than the method I used?

Cheers for your help up until now.

p.s sorry the picture is sideways, I took it upright with my phone, don't know what happened there :S

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12. Mar 12, 2014

### Simon Bridge

OK - that works ... it was pretty much what I had in mind.
Notice how labeling things makes them easier to talk/write about?

The way to get more accurate is to leave evaluating the trig functions to the end... but it looks like you only need your answers to 2 sig fig.

It gets faster with experience.