Check out this attachment on vector Statics..
vectorsstatics.pdf
I'm trying to calculate the components of an applied load at an angle and am struggling to make sense of it.
For instance a load applied at a 45 degree angle to a wall breaks down into 50% of load going in Z (up and down) and 50% in lets say X (horizontal). So for 100 lbsf that would be 50 lbsf and 50 lbsf. But when you run the math using the trig sin45 = O/H...or sin45 = O/100... I get around 70lbsf and 70lbsf as sin45 and cos45 is .707.
I'm stumped, I know this should be easier than this. I think the trig and angles are whats confusing me and somehow I need to break down the components differently.
Appreciate your help.
Check out this attachment on vector Statics..
vectorsstatics.pdf
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Thanks for the input, I reviewed the pdf you posted but still a little lost with the application I'm looking at. I attached a pic of what I'm trying to determine and its the blue arrows that are acting on the rigid wedges from the 100 N force pulling down. I put the A on there to show that the angles are equal on the wedges, lets say A=75 degrees. My guess is that the blue arrow force would be 48.29, because you have a 100 N force, divided by 2 for each side, and then a 50 N normal force to the angled surface of the wedge, then cos15=Adjacent/50N, gets you to 48.29.
Does that sound right?
I know that wedges have a mechanical advantage but not sure I'm getting this right. Shouldn't the mechanical advantage yield more than 100N of force on the wedge.
There are a couple of wedge calculators on this webpage..
Machine Design Applications
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I re-analyzed it yesterday and I believe its actually 187 N, from what I can tell it makes sense as you get a mechanical advantage with wedges, and this yields about a 1.87 MA with the 15 degree angle. Can anybody confirm?
Your drawing indicated that you want the loading perpendicular to the applied loading on the wedges...
Your original calculations are for the loads perp. to the reactionary contact surfaces. Different angle... which I'm not sure what you're using.
See:
http://www.engineersedge.com/mechani...on_2_13973.htm
and
http://www.engineersedge.com/mechani...tion_13972.htm
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I want to know what the force exerted on the wedge is, the blue arrow. This force is perpendicular to the applied load but yes transfers through the angled wedges. Neglecting friction for now. Those examples seem similar to what I'm doing, just turned around cause in my case the applied load doesnt act directly on the wedge, but I think its the same concept.
Using your second link I get the same answer of roughly 187. Anybody agree/disagree?