OT - Any Math/static/engineer guys out there???

[Misfit6669]

I am having a brain fart and missing something on this problem. Not sure if Brandon is still around, he has saved my arse in the past but this is open for anybody.

http://i34.photobucket.com/albums/d135/misfit69/100_4631.jpg
http://i34.photobucket.com/albums/d135/misfit69/100_4630.jpg

I am not seeing how to find Theta or B, it seems like I need for info. If I pick a theta like 45 then I am fine to figure out B to be 692 lb which is a valid answer to get a mag of 1200 lb.

BOB (back of book) has an answer of 744 lb and a theta of 23.8 but I don't know how they got that without having more info. If I use 23.8 as theta then I also get 744 but again, not sure how they got it. I keep thinking that they just want me to pick an angle and find B but none of the other questions are like that. I have done the 10 problems before and after this one and got the right answers but I am stupid on this one.

WHAT AM I MISSING???

Any thoughts would be appreciated big time.

[artful-dodger]

This is funny...if anybody else had asked I'd have suggested they contact you for help! But I'll give the general approach a shot (although I haven't tried to play this particular card trick in several decades. I know I've forgotten the trig...you're the math guy here, right?)

The 600 lb. pulling force can be considered as two components (vectors)...one horizontal and one vertical.

In order to pull the load straight up, the horizontal vectors on both ropes must be equal.

The 600 lb. force from the first rope at 60 degrees will provide a vertical vector force of {insert appropriate trig function for 60 degrees here}--let's pretend it was a linear function and provided 400 vertical and 200 horizontal. so the second rope will have to supply 1200 minus 400, or 800 pounds of lift.

Once you know the vertical force (800) and horizontal force (200) needed by the second rope, a quick trip back to the trig calculator should provide the final answers.

[Misfit6669]

Thanks buddy, I got some help from the lounge. I was making it a lot harder than needed. I was missing the 300 on the x-axis on the left hand side. Once somebody showed me that, I had two numbers to do a Tan theta = 300/680 which got me the right answer, I think.

Hope all is going good buddy.

[Misfit6669]

Ok guys, got a few more for you


<a href="http://picasaweb.google.com/misfit69/MyDocuments/photo#5119764230443451490"><img src="http://lh4.google.com/misfit69/Rw0OcSFVRGI/AAAAAAAAKcA/UKZ0qK6BxUw/s400/64.jpg" /></a>

<a href="http://picasaweb.google.com/misfit69/MyDocuments/photo#5119764277688091762"><img src="http://lh3.google.com/misfit69/Rw0OfCFVRHI/AAAAAAAAKcI/OYuBEnn7JXM/s400/134.jpg" /></a>

[jdieck]

4-64:
Calculate the horizontal distance from the hand to the flex link using hd=10cos60* then add the 0.75 inch and multiply that by 16 lbs to get the torque in lb-inch. So:

Torque = (10cos60*+0.75) x 16 = 92 lb-in

[jdieck]

4-134:

Resulting Force = 20 + 50 + F1 + F2 = 140 kN

The torque perpendicular to Y axis (Force applied on X axis) is:
20 x 10 + 50 x 4 + F2 x 10 = 900 kN-m

a 140 kN force to exert the same torque of 900 applied over the x axis will need to be located at 900/140 = 6.429 m = x

The torque perpendicular to X axis (Force applied on y axis) is:
50 x 3 + F2 x 13 + F1 x 11 = 1020 kN-m

a 140 kN force to exert the same torque of 1020 applied over the y axis will need to be located at 1020/140 = 7.286 m = y

Result:
Resultant Force is 140 kN located at (x,y)=(6.429,7.286) meters