Recent content by cntchds

Doing the last step again: mbulletvbulletinitial+mblockvblockinitial=(mblock+mbullet)vfinal. or .012kg*v+.8kg*0m/s=.812kg*4.16m/s then v=(.812kg*4.16m/s)/.012kg=281.5m/s Different than my initial answer by a factor of ten. I probably misplaced a decimal place. Does this look correct? Thank...

This problem was given verbally during a class period, so I will set it up in my own words. A bullet (.012kg) is fired at a block (.800kg) hanging on a strong, massless string of length 1.6m. After the collision the bullet is embedded in the block. When the block is .8m above its original...

In the case of the stress formula that you have provided is there a way to know the max stress possible with a given material? I can see that for any value of t as it goes up stress decreases exponentially, and the opposite as it decreases. There must be a point at which the material fails from...

You have piqued my curiosity. Why are you using 4"x6" for the beam dimensions? The equation I was looking at was Deflection[meters]= ((Force[Newtons]*(Length[meters]/2))/(48 * Elastic Modulus[Newtons/Meter^2] * Moment of Inertia[meters^4])) * (2 Length[meters]^2) Is this the correct...

Nope, this is a problem for an actual project I'm doing on the side.

There are 14 bolts around the edges of the plate which affix it to a structure of very heavy aluminum. 7 (equidistance -- 4" apart) along each 27" side. I'm not sure what you mean by fixed vs simple. Thanks for your help.

I'm working on a project where I know that force will be applied by a bolt pulled through a hole in the center of an aluminum plate for load testing purposes (2200 lbs, actual worst case max force around 1500 with a 1.5 times safety factor), the dimensions (27"x6.25"x?), and the material (5052...