# Calculating the net force acting on an object from a diagram

• Humbleness
In summary, the conversation is about finding the net force acting on an object in a diagram using the laws of sines and cosines. The final answer requires both magnitude and direction, and the direction is found by calculating the angle above the horizontal line and counting it as "North of East". The same process is then applied to another problem, with the final answer being 15.6N in a direction of 50.4 degrees West.
Humbleness

## Homework Statement

Calculate the net force acting on the object in the following diagram:

## Homework Equations

cosine law / sine law
ONLY solving using cosine/sine laws.

## The Attempt at a Solution

First using the cosine law to find the magnitude of the net force:
c2=a2+b2-2abcosC
c2=122+152-2 x 12 x 15 (cos124)
c2=369-(-201)
c2=570
c=23.9N

Now using sine law to find the angle:
sina/12=sinb/15=sin124/23.9
b=sin-1 (sin124/23.9 x 15)
b=31 degrees

Now I just don't know in which direction am I supposed to say it is pointing towards? Because if you look at the diagram and my work, it looks as if it is almost straight East. What should I do?

#### Attachments

• diagram.jpg
13.7 KB · Views: 4,004
• my work.jpg
40.8 KB · Views: 1,761
Humbleness said:
ONLY solving using cosine/sine laws.
Are you saying that you are asked to find the direction of the resultant using the laws of sines and cosines only? If so, it looks like the resultant is pointing a certain number of degrees "North of East". What is that number? Hint: First find how big angle a is in your drawing.

Humbleness
So I found for angle a using sine law:
sina/12N = sin31/15N
a = sin-1(sin31/15 x 12)
a = 24.3 degrees

I'm not sure what to do after this step, and honestly I don't believe I "have to" use only cosine, sine laws, however that is what I've been taught in the lesson. I got the exact same answer (23.9N) for the magnitude of the net force by solving the exact same way as in this video:
Which is basically just adding the vectors to get the sum, which is the net force.
But I'm pretty sure the answer would require a direction as well, no?
Maybe it isn't asking me of direction too; if I were to perfectly re-draw the diagram, wouldn't the net force be facing directly towards East, just like that middle dotted line in the original diagram?

The exact question, word-for-word is like this:

Calculate the net force acting on the object indicated in the following diagram. Show your work.

Humbleness said:
a = 24.3 degrees
OK, and what is the angle that C makes above the horizontal (East) direction? That's the angle "North of East".
Humbleness said:
But I'm pretty sure the answer would require a direction as well, no?
Yes. Force is a vector and when you are asked to find a vector, you must supply magnitude and direction. If you were meant to supply the magnitude only, then there should be language in the problem to that effect.
Humbleness said:
... if I were to perfectly re-draw the diagram, wouldn't the net force be facing directly towards East, just like that middle dotted line in the original diagram?
No. I did the problem and the angle above the horizontal is small but not zero. I suggest that you redo the calculations to five decimals. You can always drop some in the end.

Humbleness
Clever...
(Showing in the attachment that it's true, it isn't exactly going east.

I re-did the calculations with 5 decimals, as you suggested, and got:
Angle a = 24.597
Angle b = 31.353

And I found based entirely on using my protractor (Because I don't know what equation to use to calculate the angle from the horizontal ground), that the angle comes around roughly 0.2 degrees north of east.

#### Attachments

• IMG_0430.JPG
28.1 KB · Views: 536
Draw a horizontal line starting at the tail of vector B. The angle between this line and vector B is 24o. Do you see why?

Humbleness
Ooohh, I see it now... That is why I was supposed to calculate by 5 decimals too, so that I come as close as possible to the actual result.
So by re-doing the calculations one more time, this time including everything (beyond just 5 decimals)
I get:
Angle a = 24.59825
Therefore the direction of the 23.9N net force will be 0.598 or better 0.6 degrees north of east.

Humbleness said:
Therefore the direction of the 23.9N net force will be 0.598 or better 0.6 degrees north of east.
Yup, that's what I got.

Humbleness
kuruman said:
Yup, that's what I got.

Yey!
Thank you so much for your help! Really, I appreciate it very much.

I just have one more solving net force from a diagram problem, which I already solved, and I would really appreciate confirmation if I got it correct, and if not, where I may have done wrong, if it's not too much trouble:

Same equations, using cosine and sine laws to find both the magnitude and then angle and direction of the net force.

First I subtracted 10N and 8N in order to get:
-8.0N (S) + 10.0N (S) = 2.0N (S)

Then I was able to use tip-to-tail method:

First using cosine law to find magnitude of net force:
c2=a2+b2-2abcosC
c2=4+289-2x2x17 (cos45)
c2=293-48
c=15.6N

Then angle:
sina/2N = sinb/17N = sin45/15.6N
b = sin-1 (sin45/15.6 x 17)
b = 50.4

Therefore the net force is 15.6N (S 50.4 W) West of South

#### Attachments

• Diagram1.jpg
13.2 KB · Views: 913
• mywork.jpg
36.1 KB · Views: 684
Humbleness said:
... if it's not too much trouble ...
It's not too much trouble but it's against forum rules. You can post only one homework problem per thread. I suggest that you post this separately and I am sure someone will be happy to check it for you. Sorry.

Humbleness

## 1. How do I identify the forces acting on an object from a diagram?

To identify the forces acting on an object from a diagram, first look for any arrows pointing towards or away from the object. These arrows represent the magnitude and direction of the force. The length of the arrow represents the magnitude, and the direction of the arrow represents the direction of the force.

## 2. How do I calculate the net force on an object from a diagram?

To calculate the net force on an object from a diagram, you must first add all of the forces acting on the object. Make sure to include the direction and magnitude of each force. Once you have all the forces, you can use the formula F = ma (force equals mass times acceleration) to calculate the net force.

## 3. What is the difference between balanced and unbalanced forces?

Balanced forces are forces that are equal in magnitude and opposite in direction. When balanced forces act on an object, the net force is zero and the object will remain at rest or continue to move at a constant velocity. Unbalanced forces, on the other hand, are forces that are not equal in magnitude or opposite in direction. When unbalanced forces act on an object, the net force is not zero and the object will experience an acceleration.

## 4. Can the net force on an object be negative?

Yes, the net force on an object can be negative. This means that the forces acting on the object are in opposite directions and the object will experience a deceleration or slowing down. It is important to pay attention to the direction of the net force to accurately predict the motion of an object.

## 5. What is the relationship between net force and acceleration?

The relationship between net force and acceleration is described by Newton's second law of motion, F = ma. This means that the net force acting on an object is directly proportional to the acceleration of the object. As the net force increases, the acceleration of the object also increases. Similarly, if the net force decreases, the acceleration of the object will also decrease.

• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
24
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
5K
• Introductory Physics Homework Help
Replies
15
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
18
Views
2K