What is the correct component force along the axis of the hook?

AI Thread Summary
The discussion revolves around calculating the correct component force along the axis of a hook, with a focus on the geometry and angles involved. Participants highlight the importance of accurately determining the angle of the screw eye with respect to the horizontal, clarifying that it should be approximately 37 degrees, not 53 degrees as initially thought. A miscalculation in the coordinate system rotation was identified as a source of error in the force calculation. After correcting the geometry, the final component force was determined to be 147 lbs, confirming the accuracy of the revised approach. The conversation emphasizes the critical nature of precise angle measurements in solving such problems.
jgreen520
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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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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.
 
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.
 
jgreen520 said:
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.
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.
 
Yes the 4v and 3h is given in the problem. Inverse Tan gives you 53 degrees from the horizontal.
 
jgreen520 said:
Yes the 4v and 3h is given in the problem. Inverse Tan gives you 53 degrees from the horizontal.
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.
 
...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!
 
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