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

AI Thread Summary
The discussion focuses on calculating the net force acting on an object using vector components. The original poster is confused about how to incorporate three forces (17N, 10N, and 8N) and their angles into the calculations. They initially attempted to split the problem into two parts, but received feedback indicating that their approach was flawed, particularly in counting the 17N force twice. A more effective method suggested involves calculating the components of each force in the east-west and north-south directions, followed by vector addition. The discussion emphasizes the importance of understanding vector components and proper addition techniques to arrive at the correct net force.
barroncutter
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Homework Statement


diagrams-png.59601.png


This question is just for b)

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


Homework Equations



√ F12 + F22 = net force

c2 = a2 + b2 -2ab cos A[/B]

The Attempt at a Solution


[/B]
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 c2C2 = 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.
 
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barroncutter said:

The Attempt at a Solution


[/B]
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 c2C2 = 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.

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

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).
 
barroncutter said:
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).

And what's your answer?
 
PeroK said:
And what's your answer?

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 c2C2 = 102 + 172 – 2(10)(17)cos45° = 33.6 N
 
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!
 
PeroK said:
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.

Alright, thank you for your answer. Do you mind showing me how to do it this way?
 
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).
 
The most obvious first step is to add the 10N and 8N forces together. Then you only have two forces to deal with.
 

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