# Calculate the Net Force

1. May 26, 2015

### MaryBarnes

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

2. Relevant equations

3. The attempt at a solution
38)
c^2=a^2=b^2-2abcosC
c=([32.0N]^2=[38.0N]^2-2[32N]*[38N]cos125)1/2
c=62.15N

sin law
32/sinA=62.2N/sin125
=24.8 or 25

2. May 26, 2015

### SteamKing

Staff Emeritus
Unless I misunderstand the problem statement, the net force here will be the algebraic sum of the two force vectors. Applying the law of cosines as you did gives you the difference of the two force vectors.

Also, you must supply a direction along with the magnitude of the force in order to have a complete answer.

3. May 27, 2015

### helium4

As SteamKing mentioned, you've calculated the magnitude of the difference of the two vectors, not the magnitude of the net force, which you would get by a vector sum.

The point of all these problems is to resolve the forces into two perpendicular components, an x-component and y-component, by using trigonometry, in order to simplify the addition of vectors. After you've resolved all the forces into their components, you just add all the x-components together to get a net force $F_x$ in the x-direction and add all the y-components to get a net force $F_y$ in the y-direction. The total net force $F_{net}$ will be the vector sum of $F_x$ and $F_y$, and since these two are perpendicular to each other, you just need to use the Pythagorean theorem to determine that the magnitude of $F_{net}$ will be $\sqrt{F_x^2+F_y^2}$. To determine the direction of the net force, you need to find the angle in the triangle with sides $F_x$, $F_y$ and $F_{net}$, which can be determined using $\tan{\alpha}=\frac{F_y}{F_x}$, and thus $\alpha=\arctan{\frac{F_y}{F_x}}$.

You can find some examples at http://www.physicsclassroom.com/Class/vectors/u3l3b.cfm and http://www.physicsclassroom.com/class/vectors/Lesson-3/Net-Force-Problems-Revisited.

4. May 27, 2015

### MaryBarnes

I left the vectors out of the equation, but i do know that i need to put them into my final answer.

we haven't been shown how to break things down into components in this unit, so we wouldn't be required to use it. The only examples shown to us how to find net force is this formula: c^2=a^2+b^2 -2abcosC would this be the right formula for finding the resultant force?

Last edited: May 27, 2015
5. May 27, 2015

### MaryBarnes

i need to find the angle in order to figure out the direction. what formula do i use?

6. May 27, 2015

### MaryBarnes

Also, can anyone help me with the other two diagrams? I uploaded them in my initial post.

7. May 27, 2015

### SteamKing

Staff Emeritus
You've been given some other hints by Helium4. Have you tried to use those to solve this problem?

BTW, concentrate on solving this problem first. The other problems are similar, and you won't be able to tackle them successfully if you don't understand what's going on here.

8. May 27, 2015

### MaryBarnes

The way that it is shown to me is using sinA/A=sinB/B=sinC/C or tan=opp/adj

so what i did was 62.2N/sin125=32.0N/sin
=24.8 or 25 deg

9. May 27, 2015

### helium4

I'm assuming you haven't learned about force components, so a different solution to this problem might be to complete a parallelogram with sides equal to the two force vectors and use a Cosine Rule in that parallelogram (see the attached image). When you have completed such a parallelogram, you can use the Cosine Rule in the triangle with sides $F_1$, $F_2$ and $F_{net}$, as shown in the image, from which you can determine $F_{net}$ using $$F_{net}^2=F_1^2+F_2^2-2\cdot F_1 \cdot F_2 \cdot \cos{(180 -\alpha)}$$ Once you've done that, then you can also find the angle of $F_{net}$ using the Sine Rule in the same triangle.

So, try completing a parallelogram in your problem and see if you can apply the Cosine Rule to that parallelogram.

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Last edited: May 27, 2015
10. May 27, 2015

### MaryBarnes

this is what i get by using the method you suggested to me:

180deg-35deg=145deg

Fnet^2= F1^2+F^2 -2(F1)(F2) cosC
Fnet^2=32N^2+38^2 -2(32N)(38N)cos145
Fnet^2= 4460.17
square root of 4460.17 = 66.78 or 67N

sin 145/67N=sin/38
sin=38N sin145/67N
sin=21.79/67N
sin-1(0.3253)
=18.9 or 19deg

Fnet=67N [N 19deg W]

is this right?

11. May 27, 2015

### helium4

This is still incorrect because the case I described holds for a general case where $\alpha < 90$, which isn't the case in your problem #38, but holds for example for the last problem you attached. For the case where $\alpha > 90$, which is the case in problem #38, you would need to draw a slightly different parallelogram, but the principle is the same: you want the diagonal to be $F_{net}$ and the two sides to be the two given vectors $F_1$ and $F_2$, from which you can determine $F_{net}$ using the Cosine Rule in that triangle. To help you a bit more, see the following drawing for problem #38, where I've calculated the angle between $F_1=38N$ and $F_2=32N$ to be 180-90-35=55 degrees. Now you should be able to calculate $F_{net}$ using Cosine Rule and its angle using the Sine Rule.

Last edited: May 27, 2015
12. May 27, 2015

### MaryBarnes

I knew that the angle part couldn't be right!

so this would be my answer now that i have changed the angle to 55deg:

90deg-35=55deg

Fnet^2= F1^2+F^2 -2(F1)(F2) cosC
Fnet^2=32N^2+38^2 -2(32N)(38N)cos55
Fnet^2= 1073
square root of 1073 = 32.7N or 33N

sin 55/33N=sin/38
sin=38N sin55/33N
sin=31.12/33
sin-1(0.9432)
=70.6deg or 71 deg

Fnet= 33N [N 71deg W]

if this is correct, i think i should be able to answer the second and third question

13. May 27, 2015

### helium4

Great, you're almost there. The magnitude of the net force is correct, but you calculated a different angle. If you look at the drawing, you'll see that the angle you calculated is actually the angle between the vectors $F_{net}=33N$ and $F_2=32N$, but what you want is the angle between the vector $F_{net}$ and the horizontal. So, what would be the correct angle?

Last edited: May 27, 2015
14. May 27, 2015

### MaryBarnes

angles are my weakness! would i do sin55/33=sin/32 ?

15. May 27, 2015

### SteamKing

Staff Emeritus
You need an argument for the sin/32 bit. IOW, which angle are you talking about?

16. May 27, 2015

### helium4

Well, you have more options. First, you can use the angle 71° between $F_{net}=33N$ and $F_2=32N$ that you've already calculated and use the fact that the angle $\alpha$ (angle between $F_{net}$ and the horizontal), the 71° angle and a 90° angle are in the same triangle, and so you can determine $\alpha$ from that triangle.

Another option is to find the angle $\beta$ between $F_{net}$ and $F_1=38N$, which is what you tried to do now with the Sine Rule $\frac{sin{55}}{33}=\frac{\sin{\beta}}{32}$. For this angle, you know that $\beta=\alpha+35°$ (see the drawing), and so you can also determine $\alpha$ from there.

Also, as SteamKing pointed out, it would be nice to write the sine of angle $\theta$ as $\sin{\theta}$ instead of just $\sin$, since we don't know what angle you're talking about.

So, what's the final angle $\alpha$, then?

Last edited: May 27, 2015
17. May 27, 2015

### MaryBarnes

I don't know how to write down the symbols for the angles, how do you do that?

anyways, after doing sin 55/33=sin/32 I got 52.59deg. is this right?

18. May 27, 2015

### MaryBarnes

10N=8.0N[N]
10N-8.0=2N

Fnet^2= F1^2+F^2 -2(F1)(F2) cosC
Fnet^2=2n^2+17n^2-2*2n*17Ncos45
Fnet^2=244.9
square root of 244.9=15.6N or 16N
sin/17N=sin45/15.6
=12/15.6
sin-1(o.77)
=50deg
50deg-90deg=40deg

16N [W 40deg S]

19. May 27, 2015

### MaryBarnes

180deg-56deg=124
Fnet^2= F1^2+F^2 -2(F1)(F2) cosC
Fnet^2=15^2=12N^2-2(15N)(12N)cos124
=24N

sin/12=sin56/24
=sin-1(0.4145)
=24.5

24N [W 24deg S]

20. May 28, 2015

### helium4

I think we were talking about different angles. The angle between the two force vectors (55°) you calculated earlier was correct and the magnitude of the net force (33N) was correct:
But the direction of the net force (i.e. the angle $\alpha$ between $F_{net}$ and the horizontal) was incorrect, because if you look at the attached image, the 71° angle you calculated is between the net force vector and the 32N force vector, not between $F_{net}$ and the horizontal:
What you want is the angle $\alpha$, since that gives you the direction of $F_{net}$. Now it should be pretty easy to calculate $\alpha$ from the image.