# Calculating net force from a diagram

• pbonnie
In summary, the homework statement is trying to calculate the net force acting on each object in the following diagrams, but is having trouble with the angle between the resultant vector and the horizontal.
pbonnie

## Homework Statement

Calculate the net force acting on each object indicated in the following diagrams.

## Homework Equations

c^2 = a^2 + b^2
sinA/a = sinB/b = sinC/c
c^2 = a^2 + b^2 - 2abcosC

## The Attempt at a Solution

Fnet =?
Fnet(v) = 8.0 N [N] + 10.0 N
= 10.0 N - 8.0 N [N]
=2.0 N

I'm not sure what to do after this.
I know the angle opposite the 45° is 180° - 45° = 135°
But I'm not sure how to add the 17 N vector and the 2.0 N vector.

#### Attachments

• picture.jpg
2.2 KB · Views: 521

Welcome to PF pbonnie!

That diagram is really tiny!

Anyway, remember that vectors add "tip to tail" so you slide the 2.0 N southward vector over until its "tail" touches the "tip" of the 17.0 N northwest-ward vector. The resultant then goes from the tail of the 17.0 N to the tip of the 2.0 N. So these three vectors form a triangle, in which you know two of the sides and one angle. This should be enough to apply the cosine law, right?

Thank you for the quick reply!
Sorry for the tiny picture.

Does the 135° angle sit at the corner of 2.0 N and 17.0 N vector?
So like this? (picture attached, hopefully normal sized!)

#### Attachments

• angle.jpg
3.3 KB · Views: 441

pbonnie said:
Thank you for the quick reply!
Sorry for the tiny picture.

Does the 135° angle sit at the corner of 2.0 N and 17.0 N vector?
So like this? (picture attached, hopefully normal sized!)

Well, yeah, sort of. The angle at that vertex of the triangle is 45 degrees. What you have is 180 minus that, but we want the interior, acute angle, not the exterior one.

Another way to think about it is that the 17.0 N vector is angled 45 degrees from the vertical, and the 2.0 N vector is vertical. Hence the angle between them is...45 degrees.

Ohh okay. I thought that didn't seem right.
So then I could do cosine law..
c^2 = 2.0N^2 + 17.0N^2 - 2(2.0N)(17.0N)cos45°
*showing work on actual sent in work*
c = 15.6 N
then find the angle of the vector using sine law, so..
sinA/2.0N = sin 45/15.6N
*show work*
A = 5°

The net force is 15.6 N [W 5° N]?

pbonnie said:
Ohh okay. I thought that didn't seem right.
So then I could do cosine law..
c^2 = 2.0N^2 + 17.0N^2 - 2(2.0N)(17.0N)cos45°
*showing work on actual sent in work*
c = 15.6 N
then find the angle of the vector using sine law, so..
sinA/2.0N = sin 45/15.6N
*show work*
A = 5°

The net force is 15.6 N [W 5° N]?

I agree with your magnitude. Your direction is wrong. Ask yourself: WHICH angle are you measuring when you use the sine law like that? You're measuring the one IN the triangle. But this is the angle between the resultant vector and the 17.0 N vector. What you want is the angle between the resultant vector and the horizontal.

Hint: the 17.0 N vector is 45 deg. from the horizontal. You know the angle between the 17.0 N vector and the resultant. Therefore, you know the angle between the resultant and the horizontal.

Hmm.. I don't think I understand. What is the horizontal?

pbonnie said:
Hmm.. I don't think I understand. What is the horizontal?

I'm calling the east-west direction "horizontal" and the north-south direction "vertical", because this is the way they are oriented on your piece of paper.

Horizontal. Flat: ----------------------------------

So you can't say that the angle you got is "north of west" because you're not measuring it relative to due west. You should be. Draw a picture.

So the resultant is also 45 degree from the horizontal? I drew a picture but I don't know how to figure out the angle.

pbonnie said:
So the resultant is also 45 degree from the horizontal? I drew a picture but I don't know how to figure out the angle.

Nope, the 17.0 N vector is 45 from the horizontal. The resultant vector is few degrees below that (as you can see from your triangle). And you know how many degrees below that, because you just calculated it.

Oh! So the angle between the resultant and the horizontal would be 5 + 45, so 50°.
Then it's 15.6 N [W 50° N]?

Or sorry, it would be 45 MINUS the 5, so 40

pbonnie said:
Or sorry, it would be 45 MINUS the 5, so 40

Correct! If you're facing due west, you need to rotate towards the north (clockwise) 45 deg to be facing in the direction of the 17.0 N vector. But if you were trying to find the direction of the resultant vector, then you've gone too far, and you need to rotate back (counterclockwise) by 5 deg. Hence the answer is 45 - 5.

Thank you very much!

## 1. How do you calculate the net force from a diagram?

The net force from a diagram can be calculated by adding up all the individual forces acting on an object. This can be done by using vector addition, where the direction and magnitude of each force is taken into account.

## 2. What information do I need from the diagram to calculate the net force?

You will need to know the magnitude and direction of each individual force acting on the object, as well as the angle at which each force is acting. This information can be represented using arrows and labeled in the diagram.

## 3. Can the net force ever be zero?

Yes, the net force can be zero if all the individual forces acting on the object cancel each other out. This can happen if the forces are equal in magnitude but opposite in direction.

## 4. How do I determine the direction of the net force?

The direction of the net force can be determined by looking at the direction in which the object will accelerate. If the net force is in the same direction as the object's motion, it will accelerate in that direction. If the net force is in the opposite direction, it will decelerate or move in the opposite direction.

## 5. Is it possible to have a negative net force?

Yes, it is possible to have a negative net force. This means that the object will experience a force in the opposite direction of its motion, causing it to decelerate or move in the opposite direction.

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