Calculating the net force acting on an object from a diagram

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  • #1
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Homework Statement


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


Homework Equations


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

The Attempt at a Solution


my work.jpg

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?
 

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Answers and Replies

  • #2
kuruman
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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.
 
  • #3
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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:
https://www.khanacademy.org/math/pr...ns-of-vectors/v/vector-component-in-direction
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.
 
  • #4
kuruman
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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".
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.
... 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.
 
  • #5
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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.
 

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  • #6
kuruman
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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?
 
  • #7
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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.
 
  • #8
kuruman
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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.
 
  • #9
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Yup, that's what I got.

Yey! :smile:
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:
Diagram1.jpg


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:
mywork.jpg


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
 

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  • #10
kuruman
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... 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.
 

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