Question

Yelp me please!!!!

Answer 1

I've already done the question but can someone please check it for me!!! ?!?!?! d_p

Answer 2

@Astrophysics Please help me.....!!! :")

Answer 3

can you zoo in on it?!?!?!?!

Answer 4

@johnweldon1993 Please help me!!!

Answer 5

Just 1 sec finishing up a previous post :)

Answer 6

Okay back sorry about that...let me take a look

Answer 7

can you see my answer key?!?!?!?!?!?!

Answer 8

It didnt load at first for some reason but yeah I got it

Answer 9

So everything looks perfect...did you just want confirmation or did this turn out incorrect?

Answer 10

no just for confirmation but can you help me with net force?!?!?!?! ?

Answer 11

i know that the net force is 24N but at what angle tho?!?!?!

Answer 12

Oh okay... well to make your life SO much easier I'm going to show you a quicker easier way to do this question

And it will answer the angle question as well :)

Answer 13

Thank you!!! :")

Answer 14

i think i'm looking for the resultant angle between "C" and "A"....i think...

Answer 15

Net force of these 2 forces



If we figure out the components of these vectors...we can add them and find the resultant that way...since we know \(\large \text{Resultant} = \sqrt{F_x^2 + F_y^2}\)

So...we know that the Force and the COSINE of the angle, give us our 'x' components for the vectors...
And the Force and the SINE of the angle give us the 'y' components

*Let the 12N force be V1 and 15N be V2 and let us break them into components

\[\large V_1 = <12cos(32), 12sin(32)>\]
\[\large V_2 = <15cos(24), 15sin(24)>\]

Summing the components of the 2 Give us

\[\large \text{Resultant} = <12cos(32) + 15cos(24), 12sin(32) + 15sin(24)> \\ = <23.87976, 0.25798>\]

And the magnitude of the resultant is our Resultant vector

\[\large V = \sqrt{23.87976^2 + 0.25798^2} = 23.88N\]

Answer 16

yes i get about the same amount of force.....

Answer 17

Now I'm not sure if you've seen the whole breaking vectors into components...but it is very easy as you can see...and quick!

And 1 good thing about it is....you see how we found our resultant to be

\[\large V = <23.87976, 0.25798>\]

Well...those are the COMPONENTS of the vector

SO, a neat little trick is...to find the angle this vector is pointing...we simply take the 'y' components....divide it by the 'x' component...and take the inverse tangent of that answer

So

\[\large tan^{-1}\theta = \frac{y}{x}\]
\[\large \theta = tan^{-1} (\frac{0.25798}{23.87976}) = ??\]

Answer 18

but the thing is that all the angles are given.....so what are we looking for?!?!?!?!

Answer 19

We are to find that "single force" and that "SOME angle"

Answer 20

We already found the force...now we need to find what angle it is being applied at

Answer 21

with respect to what?

Answer 22

With respect to the horizontal

Answer 23

ok so am i using the same varialbes i used previously to find hte net force....A and C?

Answer 24

Nope those are out of the question now because we replaced those 2 forces with this single force

We are just using the components of the vector we found like I wrote above

Answer 25

ok

Answer 26

I fell like this isnt making sense...

So we have 2 forces acting on something

By adding those 2...we found a "resultant" force that we can replace those 2 forces with

i.e....it wouldnt matter if we used those 2 forces acting at those different angles....or used this 1 new vector acting at some angle....the box would feel the same exact force

Answer 27

So instead of having 2 forces of 12N acting at 32 degrees above the horizontal....and 15N acting at 24 degrees below the horizontal

We instead us just 1 force of 23.88N acting at SOME angle with respect to the horizontal

Answer 28

And that SOME angle....is solved using that formula I wrote above

\[\large \text{Angle} = \tan^{-1}(\frac{0.25798}{23.87976}) = ?\]

Answer 29

ok thank you very much.....it's too noisey here....:"D i can't focus.

Answer 30

but i think i got it.....!!! :D

Answer 31

i'm getting 0.61 deg....that's not even a one complete degree....

Answer 32

i don't want to use the component method.... x comp and y comp, please just tell me from what i've done, how do i go about find the angle....Please help :")