# Am I doing this right? (Physics, Newtons, Diagrams)

1. Jan 27, 2017

### barroncutter

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

This question is just for b)

"Calculate the force acting on the object indicated in the diagram."

2. Relevant equations

√ F12 + F22 = net force

c2 = a2 + b2 -2ab cos A

3. The attempt at a solution

So I first started by using a technique my friend taught me to find the force when there isn't an angle greater than 90 degrees. So I split the question into two parts. The upper part with the 45 degree angle, and the net force where the larger angle was.

First Net Force

√ F12 + F22 = net force (north, west)

F1 = 8 + 17Ncos45° = 20.02

F2 = 17Ncos45° = 12.02

√ 545.28 = 23.4 N

Second Net Force

Angle between x-axis & 17N: 90° - 45° = 45°

This angle will always give you the angle you need to solve c2

C2 = 102 + 172 – 2(10)(17)cos45° = 33.6 N

Now I'm assuming I need to add both of these together? I'm also a little confused on how I'm going to find the angle of them. Because how do I add both net forces and their angles? A little confused by this.

Last edited: Jan 27, 2017
2. Jan 27, 2017

### PeroK

Well, I'm confused by what you're doing. Question b) you have three forces? 17N, 10N and 8N?

3. Jan 27, 2017

### barroncutter

Yes. The question was "Calculate the force acting on the object indicated in the diagram.". To be honest, I'm a little confused because of that third force (17N).

4. Jan 27, 2017

### PeroK

5. Jan 27, 2017

### barroncutter

So I tried to split the answer in two. Using 17N and 8N to find the net force between these two, then in the second part of the equation I'm trying to find the force between 17N and 10N. Hence why I did this:

17N & 8N Net Force:
√ F12 + F22 = net force (north, west)

F1 = 8 + 17Ncos45° = 20.02

F2 = 17Ncos45° = 12.02

√ 545.28 = 23.4 N

17N & 10N Net Force:

Angle between x-axis & 17N: 90° - 45° = 45°

This angle will always give you the angle you need to solve c2

C2 = 102 + 172 – 2(10)(17)cos45° = 33.6 N

6. Jan 27, 2017

### PeroK

If your answer is 33.6N, then you must see that is impossible. The maximum it could possibly be is 35N and that's if all the forces were in a straight line. And, in this case the 8N and 10N forces are in opposite directions and must largely cancel each other out.

I don't understand at all how you are trying to get an answer. I can't advise you on how to fix things, because it makes no sense to me.

Alternatively, any force is completely determined by its components in each direction. So, a simpler and more reliable approach is to calculate the overall components in the two directions: EW and NS.

You may want to try doing the problem that way.

PS another approach is to draw a vector diagram. That is good for getting a quick view of the force, but you'll still need to do the calculations.

PPS: Okay, I see, your counting the 17N force twice!

7. Jan 27, 2017

### barroncutter

Alright, thank you for your answer. Do you mind showing me how to do it this way?

8. Jan 27, 2017

### PeroK

You have three forces. Each has an EW component and a NS component. Two of the forces are very simple as they are in the NS direction. I can see that you do know how to calculate the components of a vector using $cos$ and $sin$. You need to apply that to the third force to get its components.

Then you add the components together. This is an example of vector addition, if you know what that is.

You also need a $\pm$ convention. The normal one is that right is +ve, left is -ve (in the EW direction); and, up is +ve and down -ve (in the NS direction).

9. Jan 28, 2017

### CWatters

The most obvious first step is to add the 10N and 8N forces together. Then you only have two forces to deal with.