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Eminence · Oct 30, 2010, 06:26 PM

Hi, I having been trying to work out this problem algebraicly but am having no luck, there are a few things I am not sure about.
I have uploaded an image I drew in paint so I hope it shows up. What I'm not sure about is when I work out Cos/Sin Would I use the degress from the horizontal or what?

IF the image didn't appear, There's 4 Forces the first is 30 degrees at 10 N from the horizontal, 2nd is 120 degrees, 10N, 3rd is 200 degrees , 5N, and the 4th is 325 degrees, 10N

Here's a link for the images on fileshack http://img709.imageshack.us/i/65081559.jpg/

Unknown008 · Oct 30, 2010, 11:16 PM

Here's your image. You need to put the [img] tags between the link and use the links on the right side of the window, under 'Links to share this image' and 'Direct'.

I would advise working out the vertical and horizontal components.

Could you show us what you have tried so far?

Eminence · Oct 30, 2010, 11:19 PM

Yeah, I tried working out the vertical/horizontal components but I worked them out from the closest 0/90/180/270 point which I'm very sure gave me the wrong answer and I'm not sure if have to start all forces at the 0 point

Unknown008 · Oct 30, 2010, 11:24 PM

Yes, you work them from those points indeed.

Taking all the horizontal components and positive direction being to the right (I'm taking each of the forces in an anticlockwise direction from the horizontal):

10cos(30) - 10 cos(60) - 5cos(20) + 10cos(35) = ?

Taking all the vertical components and positive direction being upwards (I'm taking each of the forces in an anticlockwise direction from the horizontal):

10sin(30) + 10sin(60) - 5sin(20) - 10sin(35) = ?

Can you continue now?

Here's a picture of where I got those angles, except the last force should be 10 N instead of 5 N, sorry for that.

Eminence · Oct 30, 2010, 11:54 PM

Ah I see, that makes much more sense just one more thing, is 10sin(30)+10sin(30) meant to be 10sin(30)+10sin(60) intead of 30?

Unknown008 · Oct 30, 2010, 11:55 PM

Right, another typo from me :o

Eminence · Oct 31, 2010, 12:21 AM

Also making sure that I get it right,do I just add Fx and Fy and also what do I have to do to work out at what degrees it is at?

Unknown008 · Oct 31, 2010, 12:27 AM

No, you don't directly add Fx and Fy, you use Pythagoras' Theorem.

$F_{net} = \sqrt{(F_x)^2 + (F_y)^2}$

Post the answers that you got for Fx and Fy. Depending on the direction of those two, the angle will be different, and I'm also ensuring that you got the right values. :)

Eminence · Oct 31, 2010, 12:33 AM

I got Fx=7.16 and Fy=17.68
For pythagoras I got 19.07

Unknown008 · Oct 31, 2010, 12:40 AM

Okay, close to my answers:

Fx = 7.15 N
Fy = 17.7 N

Fnet = 19.1 N

Okay, since we took up and to the right positive directions, and both Fx and Fy are positive, it means the block will move in a diagonal fashion, up to the right.

You can now make a simple sketch to show this new situation.

The angle, you will see will be obtained by:

$\frac{F_y}{F_x} = \tan\theta$

Where theta is the angle with the horizontal.

$\theta = \tan^{-1}$\frac{F_y}{F_x}$$

Eminence · Oct 31, 2010, 12:49 AM

I got 1.18 for Atan

Unknown008 · Oct 31, 2010, 12:51 AM

Well, if you are giving the answer in radians yes... but in degrees, it makes 68.0 degrees

Eminence · Oct 31, 2010, 02:56 AM

I put the angles and forces into a vector calculator http://www.1728.com/vectors.htm and also drew the angles and got roughly 9.5N at 40.9 degress which different from what we did, which makes me think that it was done wrong

Unknown008 · Oct 31, 2010, 03:09 AM

Oh, duh, I added the last vector while it should have been subtracted.

This gives

Fy = 6.21 N

This makes the net force become 9.475679066 = 9.48 N and the angle 40.98224935 = 41.0 degrees