Recent content by jmarcian

hi nvn, the aluminum is 6061-T6, does the T6 mean temper? I am not sure. as for the formulas, to compute stress i used stress=(load)/(cross-sectional area) and to compute the strain i used strain=(displacement)/(orignial length, 4in). now class mates measured the thickness, width...

"toughest" material by terms of "mechanics" (physics, not the ones that work on your car). toughness is shown as the area under the curve of a stress strain diagram. from the we can say that this material offers a high ultimate strength, but also can strain a fair amount before breaking...

thanks nvn, i guess i need some reassurance. plus measuring the displacement from the negative standpoint through me off... i converted all the data to strain and stress and came up with a graph... see attached. looks acceptable to me. i was just didnt know that aluminum could stretch that much...

ok the device used is called a “Instron” electromechanical testing machine. not sure if anyone is familiar with this, but that isn't the real point. ok, so an aluminum beam, dimensions: "ext. gauge length: 4in" "Spec. gauge length: 2in" "thickness: 0.037in" "width: 0.5in" now, the...

yes, you got it now!

on which axis is the box accelerating, (or deccelerating)? does the box have any y-direction movement? think about those, and remember \sumF=ma, this means all the forces on the same axis are equal to the mass of the body times its acceleration.. breaking F=ma into components we get...

F=ma is your best friend :)

think about what andrew mason asked you... if the runner only has an x-component of velocity, is the y-component of the arrow relevant?

ok look at it this way.. PMI= if f(a), where a is some constant, is true for statement, then f(k) is true, from there if f(k+1) is true, the statement it true F(2) is what? 22-1<(2)!, 21=1<2=2! this is true, so it follows that f(k) is true! then we assume 2k-1<k! now we need to...

ok I've been thinking about this and i think it might be two questions in one. first, to analyze the linear motion of the center of the circle G. i would just treat the center as a particle, and use F=ma. draw out your FBD, use that 5-12-13 triangle instead of trig functions but in the same...

i would assume exponential growth rate here... use: P=Ae^(rt) A is initial quantity, P is quantity at time t, t is time of course, and r is rate in the first part of the equation they give you enough info to solve for r, (using ln's which i assume you know how to do), now that you know r...

i arrived at solution a also... although i can't give you a written out explanation, i can try to point you in the write direction... a good first step is to draw out that FBD... for y components you should see you have the weight of the box (downward) and the normal force on the bottom of...

seems you used the correct equations, everything looks good to me!

yes! (in between p and a!)

WHOOPS! ill keep that in mind! in your drawing, tweety, notice r is not in the appropriate place. because having the weight where you placed it would not make the bar rotate about point P, but Q instead...