Recent content by celtics777

Homework Statement Several of us claimed that if d=gcd(a,b,c) then d is a linear combination of a,b and c, i.e. that d=sa+tb+uc for some integers s,t, and u. That is true, but we only proved the analogous claim for the greatest common divisor of two numbers, i.e. when d=gcd(a,b). We need...

Homework Statement Argue that (17^4)*(5^10)*(3^5) is not the square of an integer. Homework Equations N/A? The Attempt at a Solution Do I break these up, and show that each is not a square? I'm not sure if that would be correct, but sqrt(17^4)=289 * sqrt(5^10)=3125 *...

Homework Statement If d=gcd(a,b) show that gcd((a/d),(b/d))=1 Homework Equations N/A? The Attempt at a Solution Basically, I know that I need to show that 1 is a linear combination of a/d and b/d. I'm not exactly sure how to go about this. Dividing by d gives...

Ok. I remember the right hand rule now. With my problem does this mean (-68.952 lb-in)+(68.894 lb-in) = -.058 lb-in cw?

Oh, that seems fairly odd. Where ccw is counterclockwise and cw is clockwise?

Oooh I see. That makes sense now. What do I do now that I know pushing down is ccw and to the right is cw? Does this mean cw is pos and ccw is neg?

If I push down it would rotate cw. I'm having a harder time picturing what would happen if I push "to the right" would it also be cw?

The original picture right? The rod would rotate clockwise.

I'm familiar with the cross product, but not sure how to use it to determine moment.

i'm not sure how to tell if it would be cw or ccw.

I attached an update to my work. Am I doing this right? Where should I go from here?

Homework Statement A 20-lb force is applied to the control rod Ab as shown. Knowing that the length of the rod is 9 in. and that α=25°, determine the moment of the force about point B by resolving the force into horizontal and vertical components. The first image I provided was given...

M=Fd Vertical: M=(20 lb)(9")(sin(25)) approx. 76.07 lb-in and Horizontal: M=(20 lb)(9")(cos(25)) approx. 163.14 lb-in Are those the two separate moments? Do I just add them? And what should the diagram look like? Just a right angle with 20cos(25) horizontal and 20sin(25) vertical? Thanks...

haha. I'm not drunk. I'm trying. 20sin(25)? Approx. 8.452 lb?

I'm trying really hard over here, but I'm having trouble knowing what I should be looking at on my image. I know both AB and the force already displayed is 20-lb, but which do I make the right angles to? Or am I not supposed to do that here? The one with alpha makes most sense, but then why...