# Statics - component force

1. Jan 19, 2013

### jgreen520

I worked out the attached problem. I was just looking for a logic check on my work. The problem asks you to find the component force along the axis of the hook.

Thanks

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2. Jan 19, 2013

### PhanthomJay

Your geometry and trig is off...recheck the angle. If the incline slopes at 4v:3h, then the screw eye slopes at ???
Then properly determine the angle of the screw eye with the horizontal. Your calc for Fy would them be ok, as long as you realize that Fy lies along the axis of the screw eye which is the y axis, not vertically as you have shown.

3. Jan 19, 2013

### jgreen520

Yes you are correct, the angle is supposed to be 53.13 degrees from the horizontal. That makes the angle between the resultant force (150 lbs) and the axis of the eye 28.13 degrees.

However my Fy calculation now seems to give me the wrong answer...I rotated the coordinate system which I think is giving me the problem.

4. Jan 19, 2013

### PhanthomJay

you still have the wrong angle. It is given that the slope of the incline is 4v:3h, is that correct? If so, the slope of the screw eye is not the same as the slope of the incline.

5. Jan 19, 2013

### jgreen520

Yes the 4v and 3h is given in the problem. Inverse Tan gives you 53 degrees from the horizontal.

6. Jan 19, 2013

### PhanthomJay

What I am trying to point out is that the screw eye does not make an angle of 53 degrees with the horizontal, it makes an angle of 53 degrees with the vertical. So the angle it makes with the horizontal is about 37 degrees. Redo the geometry.

7. Jan 20, 2013

### jgreen520

...I found the mistake I was making. Just needed to extend the axis line of the hook to the vertex of the slope triangle to get the correct angle. Which ends up giving me a component force of 147 lbs which is the correct answer.

Thanks!