# Resultant force

1. Sep 21, 2010

### 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 points

In 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

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Last edited: Sep 22, 2010
2. Sep 22, 2010

### Delphi51

Welcome to PF, pinksunbeam!
We have no idea what T6 is - diagram needed.

3. Sep 22, 2010

### 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.

4. Sep 22, 2010

### pinksunbeam

i just dont know where to begin!!!

5. Sep 22, 2010

### 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!

6. Sep 22, 2010

### 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. im not sure where they are calculating this from. and 62.8 for resultant.

7. Sep 22, 2010

### pinksunbeam

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

8. Sep 22, 2010

### Delphi51

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

9. Sep 22, 2010

### pinksunbeam

switching sine and cosine where?

10. Sep 22, 2010

### 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

11. Sep 22, 2010

### 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.

12. Sep 23, 2010

### 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

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?

13. Sep 23, 2010

### Delphi51

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