Calculate the Net Force Acting on the Object.

In summary, the question is asking to calculate the net force acting on a wagon that is being pulled by two separate forces. The forces are 12N at 32° and 15N at 24°. The equations provided are a^2 = b^2 + c^2, Sin A/a = Sin B/b = Sin C/c, and c^2 = a^2 + b^2 - 2abCosC. The attempt at a solution involved drawing a vector triangle, but it can also be solved using components.
  • #1
Lewis_44
2
0

Homework Statement


You have to calculate the net force acting on a wagon that is being pulled by two separate forces.

Roping being pulled: 12N /32°
|-----------------| /
| |------------
|-----------------| \

Rope being pulled: 15 N / 24°

Homework Equations


This question has stumped me! For other ones I had to use:

a^2 = b^2 + c^2
Sin A/a = Sin B/b = Sin C/c
c^2 = a^2 + b^2 - 2abCosC

The Attempt at a Solution


I have no idea where to start for this PF! I'm sorry! Any help would be greatly appreciated! The other questions I had were easy triangles but this I just can't seem to do.
 
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  • #2
Welcome to PF!

Hi Lewis_44! Welcome to PF! :wink:

Let's see… you've drawn a triangle with sides 12 15 and and x (unknown) …

can you describe this triangle (or copy it)?

ie which angles are where? :smile:
 
  • #3
Hi Lewis_44, You are giving Newtons and degrees
since this is acting on a train, I suppose you are looking for the net force 'along the path of the train' that would be the sum of the projections of the forces for which you have the magnitude and the angle with respect to said path.
Try to find those projections, this is just basic trigo, and add the values

Cheers...
 
  • #4


tiny-tim said:
Hi Lewis_44! Welcome to PF! :wink:

Let's see… you've drawn a triangle with sides 12 15 and and x (unknown) …

can you describe this triangle (or copy it)?

ie which angles are where? :smile:



Thank you so much for trying to help me! :D I've attached a picture on paint
 

Attachments

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  • #5
Lewis_44 said:
I've attached a picture on paint

ah, no …

to apply those equations (in your question), you need to draw a vector triangle :wink:

(if you don't know what that is, look it up in the PF Library or in wikipedia)
 
  • #6
tiny-tim said:
ah, no …

to apply those equations (in your question), you need to draw a vector triangle :wink:

(if you don't know what that is, look it up in the PF Library or in wikipedia)

Hello Tiny-Tim!

I have a question that's kind of like this... do I have to use the vector triangle as well?
 

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  • Similar Problem.png
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  • #7
Hi Tiny-tim
It seems to me that his drawing confirms what I was supposing.
There is no need to make vector aditions, I believe this is a simple problem of getting the net force 'along the rails', so just adding the two magnitudes time their respective cosines should do the trick.
Do I miss something ?

Cheers...
 
  • #8
Welcome to PF!

Hello TheGibby13! Welcome to PF! :smile:

You never have to use a vector triangle, you can always use components instead.

(Lewis_44 only had to if he wanted to use those trig equations)

Of course, your lines should always have arrows on them, and they should make a closed triangle! :wink:
 
  • #9
Ohh okay! Thank you tiny-tim! :D
 

1. How do you calculate net force on an object?

To calculate net force on an object, you need to add up all the forces acting on the object. This includes both the magnitude and direction of each force.

2. What is the formula for calculating net force?

The formula for calculating net force is Fnet = ΣF, which means the net force (Fnet) is equal to the sum (Σ) of all the forces (F) acting on the object.

3. What units are used to measure net force?

Net force is measured in Newtons (N) in the International System of Units (SI). In the British system, it is measured in pounds (lb).

4. How do you determine the direction of net force?

The direction of net force is determined by the direction of the larger force acting on the object. If the forces are equal in magnitude, the direction of net force will be the same as the direction of the forces.

5. Can net force be negative?

Yes, net force can be negative. This means that the forces acting on the object are in opposite directions, resulting in a net force in the opposite direction of the larger force.

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