How to Express the force as a Cartesian vector?

[JON123]

I am having trouble solving this problem, I also have it attached:

http://i160.photobucket.com/albums/t193/John123321_bucket/Capture.png

Here is a short version of how I got my answers, but it says it’s wrong:

X=500cos(30)=433N

i= 433sin(45)=306.2
j=-433cos(45)=-306.0
k=-500sin(60)=-250

If you can, please explain what I’m doing wrong, thanks.

 

[Hootenanny]

Welcome to Physics Forums.

I am having trouble solving this problem, I also have it attached:

http://i160.photobucket.com/albums/t193/John123321_bucket/Capture.png

Here is a short version of how I got my answers, but it says it’s wrong:

X=500cos(30)=433N

i= 433sin(45)=306.2
j=-433cos(45)=-306.0
k=-500sin(60)=-250

If you can, please explain what I’m doing wrong, thanks.

You can tell that the answer is wrong immediately as the highlighted line would suggest that F has a component in the negative z-direction. However, if you refer to the picture, it is clear that F acts "upwards" rather than "downwards". I would also check the angle (it shouldn't be sin(60).

P.S. Homework style questions should be posted in the Homework forums. I have asked that your thread be moved there by the mentors. Please make sure that your post you homework questions in the appropriate forum as it will help you get your questions answered more quickly.

 

[JON123]

Thanks for the reply. The force is acting in the negative direction because it's 60 degrees from the negative z-axis, meaning that it's 30 degrees bellow the x&y-axis. And I just had a typo on the third line, it was supposed to be k=-500cos(60)=-250 giving me the z-component. I did figure out the answer and it was 250i-354j-250k. I got it using 500sin(30)i-500cos(45)j-500cos(60)k. I was just wondering if this is the right way to setup the problem.

 

[Hootenanny]

Thanks for the reply. The force is acting in the negative direction because it's 60 degrees from the negative z-axis, meaning that it's 30 degrees bellow the x&y-axis. And I just had a typo on the third line, it was supposed to be k=-500cos(60)=-250 giving me the z-component. I did figure out the answer and it was 250i-354j-250k. I got it using 500sin(30)i-500cos(45)j-500cos(60)k. I was just wondering if this is the right way to setup the problem.

Apologies, I misread the figure.

 

[vela]

Thanks for the reply. The force is acting in the negative direction because it's 60 degrees from the negative z-axis, meaning that it's 30 degrees bellow the x&y-axis. And I just had a typo on the third line, it was supposed to be k=-500cos(60)=-250 giving me the z-component. I did figure out the answer and it was 250i-354j-250k. I got it using 500sin(30)i-500cos(45)j-500cos(60)k. I was just wondering if this is the right way to setup the problem.

Explain how you came up with 500 sin 30 for the x-component.

 

[JON123]

I figured out how to solve the problem, and i see what i was doing wrong. I came up with the 500cos(30) by going at the problem backwards because I had found out the answer from someone, but i see that's not the right way to get the answer. I solved it by using the formulas:
Fx=Fcos(alpha)
Fy=Fcos(beta)
Fx=Fcos(gamma)

Because i didn't have alpha for Fx, i found Fy & Fz and used the magnitude to solve for Fx. F=sqr(Fx^2+Fy^2+Fz^2) -> Fx=sqrt(500^2-Fy^2+Fz^2)