normal stress of section 1 = 1590N / 0.15m^2 = 10600 N/m^2
normal stress of section 2 = 1590N / 0.05m^2 = 31800 N/m^2
Could it be that easy? Feel like I'm missing something still
Area of a rectangle is length * width
So at section 1, the area A would be 30mm*5mm = 150mm = 0.15m ?
and
Section 2, the area A would be 10mm*5mm = 50mm = 0.05m ?
Then I would just plug those areas into the normal stress = P/A equation for both sections?
Thanks for the quick reply
Homework Statement
"The axial load for a given test sample carries 1590 N. Calculate tensile stress at sections (1) and (2) assuming the sample thickness is 5mm. (Rectangular cross section).
http://img208.imageshack.us/img208/121/tensilestressby1.th.jpg [Broken]
Homework Equations...
Ah excellent! Thank you very much. The equations for Ra and Rb were a little more complicated but I got the answers the back of the book has.
I'm surprised I didn't see that I was cancelling out both A's when they clearly aren't equal to each other.
Thank you very much. This forum has...
I didn't assume the cross-sectional areas were the same. The fat piece is 6 in^2 and the skinnier piece is 3 in^2.
In my thinking:
Stress of AC should = 44.44 kips/6 in^2 = 7.41
and
Stress of BC should = 55.56 kips/3 in^2 = 18.52
However those are not the answers. In the back of the...
I've been having difficulty with this problem for the past 5 hours. I am understanding some part, but I can't get the final answer.
http://img223.imageshack.us/img223/1103/1015zz1.th.jpg [Broken]
This is what I got so far:
EFx = Ra + Rb - P = 0
Ra + Rb = P
Sab = Sac + Scb =...
Says it attached and I can access it by clicking on it, but I uploaded it elsewhere as well.
http://img265.imageshack.us/img265/5564/363no0.th.jpg [Broken]
I've been having trouble with the homework problem attached as an image.
I need to find the tension in the cable and the reactions at A (Ay and Ax). In previous homeworks, I've been able to find the reactions at similar points A by finding the moment about A around a few other points. Now...
"Qa = -Qb. Right?"
So if Qal = mc(al)d(Tf-Ti) and Qwater = mc(water)d(Tf-Ti) and Qal = -Qwater, then would this be the correct equation with only one unknown?::
mc(al)d(Tf-Ti) = -[mc(water)d(Tf-Ti)] ?
Where I know both m, both c, and both Ti ?
Then I can solve for Tf which should be...
Sorry this reply took so long, and thanks for your previous post.
For Al, Ti = 80C
For Water, Ti = 20C
So I should use this equation? Qwater = mc(water)d(Tf-Ti) where I know m, c, and Ti
Qal = mc(al)d(Tf-Ti) where I know m, c, and Ti again
But now I have two unknowns and I'm stuck...
"A piece of aluminum (specific heat 910 J/kg-C) of mass 200g at 80.0 deg Celcius is dropped into a styrofoam cup filled with 100mL of water at 20.0 deg C. What are the final temperatures of the water and the aluminum?"
This is what I got:
I figured out the amount of heat energy transferred...
Yea, I completely understand and when the test does come, I'll be prepared. I'm the kind of person who likes to work backwards with problems. If I know the solution then it will register with me on how that solution came about. Hard to understand I suppose.
But I appreciate all the help
I found this site through Google, and it's pretty much identical to my problem, or is it not?
http://zebu.uoregon.edu/~probs/mech/newt/2ndNewton4/node2.html [Broken]
Thanks Doc
Alright, then we're going with that answer. I mean, this is a college course, but it is by no means a Senior or Graduate level course. So I doubt our teacher would make anything too challenging and neglect to give us needed information. Thank you for your help though; it got me thinking more.
After searching Google for a couple minutes, I found a similiar problem. However, this problem deals with a loop di loop and at the top of the loop, the riders are upside down and feel weightless.
Using Centripetal Force;
Fc = Fg = w = mg
Fc = mv^2/r = mg
mv^2 = mgr
v^2 = gr
v =...
Nope, I didn't leave anything out. The question I wrote here is verbatim to the question on our piece of paper.
We even went as far as using equations for Satellites in Orbits because the only unknown in those equations were T, the period. But after doing all the math, the velocity of this...
Here is what we came up with. I still think we are far off base:
After drawing the vector and free-body diagram which included W, Wperp, Wparallel, FN, Angle, and Acceleration, we came up with these equations.
The problem doesn't say anything about air resistance, so we neglected it...
The question only includes a single car on the track. And we can't use energy to solve this question because this question isn't from the energy chapter.
I've solved questions like this when I needed to find tensions in TWO strings when a weight is hanging from them at an equilibrium.
With two strings; therefore, two tensions, it was easy to find.
(SUM)Fx = 0 --> F2x + (-F1x) = 0 --> F2x = F1x
and
(SUM)Fy = 0 --> F1y + F2y = w
then...
The thing is, this problem comes from a chapter that does not discuss angular acceleration. Angular acceleration is discussed 3 or 4 chapters after this one. :(
This is another problem a buddy of mine and I cannot figure out if our life depended on it. Here's the question word for word:
A roller coaster track has a hill with a circular curve of radius 20 m. Find the speed of the roller coaster at the top of this hill.
Started by drawing a picture...
I along with a buddy of mine in the same course have been trying to figure out the following question. We spent at least an hour and a half to two hours and nothing makes sense. Any help would be greatly appreciated.
Here's the question as it's written:
One way to measure the acceleration...