Magnitude and direction of Vectors using head to tail rule

[bllnsr]

Homework Statement

Homework Equations

F = \sqrt{F_x^2 + F_y^2}
tan\theta = F_y / F_x

The Attempt at a Solution

Page 1 : http://i49.tinypic.com/vfw74k.jpg
Page 2 : http://i50.tinypic.com/2qspamr.jpg
Page 3 : http://i49.tinypic.com/24zgl6x.jpg
is it correct?

 

[MrWarlock616]

I think you've got the directions wrong. The x-component of forces are in opposite directions, so you need to account for that.

You've calculated the y-component of the resultant correctly, but even there you haven't specified the downward direction which could cause you to lose marks sometimes.

Since, your calculation for x-components is wrong, the magnitude and direction of the resultant is also wrong...

 

[jedishrfu]

the first two look okay the last page #3 you used vector length value so your tan angle is wrong.

A quick way to check your solution is to graph it and take measurements for the angle and length to see that your results agree.

 

[MrWarlock616]

the first two look okay the last page #3 you used vector length value so your tan angle is wrong.

Shouldn't F_x=-6000cos(60°)+2000cos(45°)?

 

[jedishrfu]

Shouldn't F_x=-6000cos(60°)+2000cos(45°)?

Yes, I think you're right. I didn't notice that.

 

[bllnsr]

so after doing Fx=-6000cos(60°)+2000cos(45°)
resultant force is 6797.915 correct?

 

[MrWarlock616]

so after doing Fx=-6000cos(60°)+2000cos(45°)
resultant force is 6797.915 correct?

Yes. And what will be the angle made by that resultant with the positive x-axis??

 

[bllnsr]

tan\theta = f_y/F_x
76.51(i didn't use minus sign in f_y)
180-76.51= 103.48
right?

 

[MrWarlock616]

tan\theta = f_y/F_x
76.51(i didn't use minus sign in f_y)
180-76.51= 103.48
right?

Correct! :)

 

[bllnsr]

Thanks for the help