# Verification for Net Force and Vectors?

Meeeessttteeehh
Hello everyone! I was given the following three questions in my homework, and I had a lot of trouble with them, so I was hoping someone could look them over! If you could just check out my pictures and say "yeah its awesome!" or "nope... [insert helpful answer here]" that would be GREAT. Thanks so much!

#### Attachments

• 39.png
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• 40.png
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• 38.png
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## Answers and Replies

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In 38, I do not understand the second hand-drawn force diagram. You have drawn the E force as N and the N force as E, leading to the wrong angle.

In 39, you end with "N 50 E", but no applied forces have a component to the E.

I haven't checked the numerics in detail, but they all look about right.

Meeeessttteeehh
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Hello meester,

Check your ##\sin 35^\circ##
 no need. But still:

Read the guidelines and don't erase the template.

Meeeessttteeehh
In 38, I do not understand the second hand-drawn force diagram. You have drawn the E force as N and the N force as E, leading to the wrong angle.

In 39, you end with "N 50 E", but no applied forces have a component to the E.

I haven't checked the numerics in detail, but they all look about right.

Thank you! I have adjusted the trig in question 38 and now have a final answer of 32.9 N [N 3 degrees E]. I will redraw the diagram. As for 39, that was a good catch. I guess you could probably tell I did this when I was tired... Thanks again!

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N 3 degrees E
I was expecting you would just get the complement, making it N 72°E.

Meeeessttteeehh
I was expecting you would just get the complement, making it N 72°E.

could you define complement? I'm not sure I understand.

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could you define complement? I'm not sure I understand.
In your original solution you got the N and E net forces swapped around, producing a resultant 18 degrees E of N. Correcting them should give you the complementary angle (90-θ), i.e 18 degrees N of E, or 72 degrees E of N.

Meeeessttteeehh
In your original solution you got the N and E net forces swapped around, producing a resultant 18 degrees E of N. Correcting them should give you the complementary angle (90-θ), i.e 18 degrees N of E, or 72 degrees E of N.

Sorry, but I still don't understand. I've been taught that when using the decomposition method I use trig to find theta. I swapped my values so they might be correct, and used tan to determine 3.0505 as theta, and I am not sure where I have gone wrong... Here is my updated picture
.

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• 38b.png
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Science Advisor
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Sorry, but I still don't understand. I've been taught that when using the decomposition method I use trig to find theta. I swapped my values so they might be correct, and used tan to determine 3.0505 as theta, and I am not sure where I have gone wrong... Here is my updated picture
. View attachment 219236
Isn't 3.0505 just the ratio of the sides? You need to apply tan-1 to that.

Meeeessttteeehh
Meeeessttteeehh
Isn't 3.0505 just the ratio of the sides? You need to apply tan-1 to that.

OH MY GOODESS. I just got the E 18 degrees N after drawing a million more triangles (more like 8 but still), but I can't believe I forgot the tan inverse. THANK YOU SO MUCH!