- #26

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Yes, that's roughly the correct value for ##\beta##. Now you can use that to calculate ##\alpha## as you did before.so now I've tried tried what you suggested and I got 24.48 deg. Is this right?

- Thread starter MaryBarnes
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- #26

- 11

- 0

Yes, that's roughly the correct value for ##\beta##. Now you can use that to calculate ##\alpha## as you did before.so now I've tried tried what you suggested and I got 24.48 deg. Is this right?

- #27

- 27

- 0

if i subtract 24deg from 24 deg i get 0? is this really the answer?so now I've tried tried what you suggested and I got 24.48 deg. Is this right?

for the third question i'm just going to break it down into components because that way makes way more sense to me. maybe that's not the expectation, but i don't think id loose any marks for doing so.

Thank you for all your help again!

- #28

- 11

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Yes, the angle of ##F_{net}## is roughly 0° (more precisely, it's something like 0.6°, so it's not quite horizontal).if i subtract 24deg from 24 deg i get 0? is this really the answer?

for the third question i'm just going to break it down into components because that way makes way more sense to me. maybe that's not the expectation, but i don't think id loose any marks for doing so.

Thank you for all your help again!

As for the problem with the three forces, you could approach it the same way as these two problems, but you would first need to combine two of the forces into a parallelogram and than take this force and the third force and combine those two into a parallelogram, which would give you the net force. However, the method with breaking the forces down into their components is in my opinion much easier than the parallelogram method, especially if there are more forces involved. Besides, it's a good skill to have for solving more complex problems.

So, good luck with that and let me know if you need any more help.

- #29

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I tried the third answer using the sin and cosine law. Is this right?Yes, the angle of ##F_{net}## is roughly 0° (more precisely, it's something like 0.6°, so it's not quite horizontal).

As for the problem with the three forces, you could approach it the same way as these two problems, but you would first need to combine two of the forces into a parallelogram and than take this force and the third force and combine those two into a parallelogram, which would give you the net force. However, the method with breaking the forces down into their components is in my opinion much easier than the parallelogram method, especially if there are more forces involved. Besides, it's a good skill to have for solving more complex problems.

So, good luck with that and let me know if you need any more help.

fnet^2=2N^2+17N^2 -2(2N)(17N)cos45

Fnet=square root of:244.9

Fnet=15.6N or 16N

sin45/15.6N=sin/2N

sin=2Nsin45/15.6N

sin-1(0.0906)

=5.2deg

45deg-5.2deg=40

Last edited:

- #30

- 11

- 0

Yes, that's correct, but you need to say that ##F_{net}## points to the left with 40° angle between the vector and the horizontal (i.e. points in the NW direction), otherwise it may seem as though it would point to the right (NE direction), which would be incorrect.I tried the third answer using the sin and cosine law. Is this right?

fnet^2=2N^2+17N^2 -2(2N)(17N)cos45

Fnet=square root of:244.9

Fnet=15.6N or 16N

sin45/15.6N=sin/2N

sin=2Nsin45/15.6N

sin-1(0.0906)

=5.2deg

45deg-5.2deg=40

- #31

- 27

- 0

Ok, thanks for mentioning that. Thank you for all of your help! Now I actually understand how work with these equations!Yes, that's correct, but you need to say that ##F_{net}## points to the left with 40° angle between the vector and the horizontal (i.e. points in the NW direction), otherwise it may seem as though it would point to the right (NE direction), which would be incorrect.

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