Resultant Force Acting on T6 Vertebrae

[pinksunbeam]

A chiropractor applies a thrust to the spinous process of T6. The magnitude is 400 N and it is applied 90 degrees to the y-axis. While this occurs the superior disc applies a force to the T6 body of 615 N, 75 degrees to the x-axis. The inferior disc applies a force to the T6 body of 620 N, 70 degrees to the x-axis. Finally, the interspinous ligament applies a force of 32 N, 60 degrees to the y-axis. Determine the resultant force acting on the T6 vertebra, and the direction with respect to the x-axis. Note: Use only two decimal pointsIn question chiropractor applies a thrust to spinous process t6 which is in the back.
attempt 1
(68.88x68.88+ 14.09x14.09) 1/2= 70.31n
attempt: 2

400nsin 0+ 615 sin 75+ -620sin 70+- 32ncos 60= -2.25
400cos0+ -615 cos 75+ 620 cos 70+ -32n sin 60= -27.04
(2.25x2.25+ 27.04x27.04) 1/2 = 25.04N

 

[Delphi51]

Welcome to PF, pinksunbeam!
We have no idea what T6 is - diagram needed.
And we can't help you until you show your attempt.

 

[pinksunbeam]

my friends have come up with 2.90 degrees and 62 or 64 for magnitude and 3rd quadrant of cartersian coordinate system i have no idea why? i was looking for force in N and then respect to x axis.

 

[pinksunbeam]

i just don't know where to begin!

 

[Delphi51]

Yikes. Did the diagram come with the question? It doesn't say whether the 400 N is to the left or the right, and it isn't shown on the diagram. "70 degrees to the x-axis" could be any of 4 directions. And so on. There are many possibilities for the signs on these components!

Shouldn't 32nsin 60 in the vertical part be 32*cos(60)?
In the horizontal part, 620*cos(70) should be positive if the diagram is correct.
32cos 60 should be 32*sin(60) in the horizontal part.

If there is any way to get the question clarified, do it!

 

[pinksunbeam]

400n is going left to right. ( apparently) that is exactly the question that came with the diagram! everyone seems to have 25.6 for angle. I am not sure where they are calculating this from. and 62.8 for resultant.

 

[pinksunbeam]

400N is F1 going left to right on x axis. it did not copy with diagram

 

[Delphi51]

Okay, the diagram makes it pretty clear. You should have the answer with those two changes, switching sine and cosine.

 

[pinksunbeam]

switching sine and cosine where?

 

[pinksunbeam]

i tried switching them everywhere i did

615 cos 75+ -620 cos 70 +- 32 sin 60= 183.96
-615 sin 75+ 620 sin 70 + - 32 cos 60= 718.28

33841.281633841.2816+515930.16 = 549771.4416549771.4416^(1/2) = 741.47 N

 

[Delphi51]

Horizontal:
615 cos 75+ -620 cos 70 +- 32 sin 60= 183.96

Vertical:
-615 sin 75+ 620 sin 70 + - 32 cos 60= 718.28

The sin/cos is fixed. I think the -620 cos 70 in the horizontal part should be positive; you show it going to the right in the diagram.

 

[pinksunbeam]

ok i redid it
i broke everything up in x and y components.
f1= -400 going right to left along x axis
f2= 615 cos 75= 159.17
f3=620cos70=212.05
f4=-32sin60=-27..71 added together = -56.49

y axis
f1=0
f2=-615 sin 75=-594.04
f3= 620sin70=582.609
f4= -32cos 60= -16 added together = -27.43

x2+y2 and get square root= 62.80N

for angle
tan-1 fy/fx= 24.43/56.49= 25.899=25.90 degrees.
i need to draw this in relation to x-axis on diagram but not sure where?

 

[Delphi51]

The vector must begin on T6, ideally at the center of its mass.