Calculate the Net Force Acting on the Object.

  • Thread starter Lewis_44
  • Start date
  • #1
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Homework Statement


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

Answers and Replies

  • #2
tiny-tim
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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
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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
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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
 

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  • #5
tiny-tim
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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
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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  • #7
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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
tiny-tim
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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
 

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