# Calculating Net force

## The Attempt at a Solution

32° + 24° = 56°

I used the tip-to-tail method to add vectors
c = ([12 N]^2 + [15 N]^2 - 2 [12 N] x [15 N] cos124°)1/2
c = 23.9 N

SinA/12 N = Sin124° / 23.9 N (i dont know how to solve this part)

can someone please explain what im supposed to do for this question, and verify if what i've done so far is correct.
thanks :)
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I used the tip-to-tail method to add vectors
c = ([12 N]^2 + [15 N]^2 - 2 [12 N] x [15 N] cos124°)1/2
c = 23.9 N

SinA/12 N = Sin124° / 23.9 N (i dont know how to solve this part)

can someone please explain what im supposed to do for this question, and verify if what i've done so far is correct.
thanks :)
some force is pulling the cart - i would have simply resolved the forces in the direction horizontal and vertical and took the sum of horizontal and vertical components and found out the resultant force as square root of sum of the square of horizontal and vertical net forces.
will i be wrong ?

haruspex
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Looks good so far.
SinA/12 N = Sin124° / 23.9 N (i dont know how to solve this part)
Doesn't look hard. Can't you just multiply both sides so that you only have sin A on the left?
i would have simply resolved the forces in the direction horizontal and vertical
Alexandria has been taught to use the parallelogram approach. At this stage, I think it is better to stick to that.

drvrm
SinA = Sin124° / 23.9 N x (12 N)
Sin124° = 0.83
so 0.83/23.9 N x (12 N) =0.42 ???

haruspex
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SinA = Sin124° / 23.9 N x (12 N)
Sin124° = 0.83
so 0.83/23.9 N x (12 N) =0.42 ???
Yes, that looks right. What is your next step?

so im having a problem with figuring out the degrees. How does 0.42 convert to degrees?

haruspex
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so im having a problem with figuring out the degrees. How does 0.42 convert to degrees?
You have sin A = 0.42, yes? Do you not have an inverse sine function of your calculator, spreadsheet, or whatever you use?
(But I feel you should keep more significant digits than 2 or your final answer will be rather inaccurate.)

So i used the inverse sine function on my calculator , and i got 24.8 degrees
so would the direction of the force be West 24.8 derees North?

haruspex
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So i used the inverse sine function on my calculator , and i got 24.8 degrees
so would the direction of the force be West 24.8 derees North?
Look at your diagram. Which angle did you label A?

wait so did i calculate the right degrees (24.8)
im confused on how to figure out the direction, just by looking at angle a in the diagram i still cannot seem to figure out the direction,
is it East 24.8 degrees North?

haruspex
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wait so did i calculate the right degrees (24.8)
im confused on how to figure out the direction, just by looking at angle a in the diagram i still cannot seem to figure out the direction,
is it East 24.8 degrees North?
Take your triangle diagram in post #1 and add a line to the East from the left-hand corner. There are three angles you know in that diagram. What other angles can you deduce?

i also re-did this question, can someone tell me if i am correct?
40.

c2 = a2 + b2 – 2abcosC

c = (a2 + b2 – 2abcosC) ½

32 degrees + 24 degrees = 56 degrees

c = ([12 N]2 + [15 N]2 - 2 [12 N] x [15 N] cos124°) ½

c = 23.9 N

SinA/a = SinB/b = SinC/c

SinA/15 N = Sin124°/23.9 N

SinA = (15 N) Sin124° / 23.9 N

A = sin^-1 (0.52)

A = 31.3 degrees

Fnet = 23.9 N [North 31.3°East]

haruspex
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i also re-did this question, can someone tell me if i am correct?
40.

c2 = a2 + b2 – 2abcosC

c = (a2 + b2 – 2abcosC) ½
View attachment 98276

32 degrees + 24 degrees = 56 degrees

c = ([12 N]2 + [15 N]2 - 2 [12 N] x [15 N] cos124°) ½

c = 23.9 N

SinA/a = SinB/b = SinC/c

SinA/15 N = Sin124°/23.9 N

SinA = (15 N) Sin124° / 23.9 N

A = sin^-1 (0.52)

A = 31.3 degrees

Fnet = 23.9 N [North 31.3°East]
Ok until the last line. You calculated the angle A (theta in the diagram). That is not the angle between the resultant and North.

so what angle in the diagram is 31.3 supposed to represent? which angle did i calculate, i need to figure out the angle between the resultant and the 12 N vector (theta) right?

haruspex
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so what angle in the diagram is 31.3 supposed to represent? which angle did i calculate, i need to figure out the angle between the resultant and the 12 N vector (theta) right?
You did calculate theta, but when you say the resultant is N x degrees E you mean x is the angle between the resultant and N. In your diagram, you have a dashed line going N. What is the angle between that and the resultant?

is it 89 degrees?

haruspex
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is it 89 degrees?
Yes. (Well, 89.3, using the 31.3 you calculated.)

i got that by doing the following
32 - 31.3 = 0.7
90 - 0.7 = 89.3

but 89.3 is the angle between the resultant and the dashed line going north, what does that have to do with the angle between the resultant and the 12 N vector (theta)?

haruspex
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i got that by doing the following
32 - 31.3 = 0.7
90 - 0.7 = 89.3

but 89.3 is the angle between the resultant and the dashes line going north, what does that have to do with the angle between the resultant and the 12 N vector (theta)?
I don't understand your question. You just wrote out the relationship between the two in your own post: angle between resultant and N is angle between E and N (90) - angle between 12N force and E (32) + angle between resultant and 12N force (theta).

I just dont understand what angle im supposed to use in my final answer, if 89.3 is the angle between the north line and the resultant, then 31.3 is the angle between the 12 N vector and the resultant, meaning Fnet is still:
23.9 N [North 31.3°East]
am i wrong?

haruspex
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I just dont understand what angle im supposed to use in my final answer, if 89.3 is the angle between the north line and the resultant, then 31.3 is the angle between the 12 N vector and the resultant, meaning Fnet is still:
23.9 N [North 31.3°East]
am i wrong?
You do not seem to understand what [North 31.3°East] means. That would mean 31.3 degrees East of due North, i.e. that to find the direction of the resultant you would face North then rotate 31.3 degrees to the East. It is unrelated to the direction of the 12N force. Clearly the angle of the resultant is 89.3 degrees East of North.

so are you saying Fnet = 23.9 [North 89.3 degrees East]?
but for the Fnet, i only need to find theta right, and theta is 31.3 degrees?

haruspex
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so are you saying Fnet = 23.9 [North 89.3 degrees East]?
Yes (but specifying the units as Newtons).
but for the Fnet, i only need to find theta right, and theta is 31.3 degrees?
What angle you need to find depends on how you are to express the answer.

You could answer that it is 31.3 degrees further to the East than the given 12N force, but that is surely not the form the answer is required in. They will want the answer given in absolute co-ordinates, independent of any circumstantial directions in the problem.
So you should answer "89.3 degrees E of N", "N 89.3 degrees E", or just "89.3 degrees" (using the navigational standard of expressing any direction as a number of degrees clockwise from North).

In a more mathematical environment, where angles are measured anticlockwise from the positive X axis, and thinking of that as being East, you might answer 0.7 degrees.

ok got it, so the final answer is:
Fnet = 23.9 N [North 89.3 degrees East]
just a thought, how would you write the direction if you wanted to use 31.3 degrees
wouldnt it be [north 31.3 degrees east], since the 12 N vecor is further to the north and the resultant is further to the east.
if i wrote that as my answer, would it also be correct?

haruspex