Recent content by 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!

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

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...

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

After seeing the examples I see what I missed! Just factoring out the common term/coefficient. I was just missing it because the equation was a bit busy. Thanks!

So the part I am curious about is why when you have 2 F_b's or say they were x's do you just factor them out. Normally once you have each variable on opposite sides of the equation you add them. So if you had 5 different x's on one side of the equation you would just factor them out and not...

I was trying to understand why in the attached equations when they divided to get F_B alone it wasn't 2B. Thanks

Thanks for the response. So differentiating D-LCos(theta) the constant D goes to 0 and the -LCos(theta) goes to +Lsin(theta) ...then multiply that by sin(theta) in the denominator to get Lsin^2(theta)... However I was curious why nothing happens to the L...in front of the Cos. Thanks

Please see attached. I was looking for an explanation of the answer I have attached. Its been a little while and was just looking for the logic behind the differentiation shown for this problem. Its basically an optimization problem where I am looking for the minimum angle (theta) for the...