alright I see that so MPA is N/mm^2 so I don't have to convert anything so it would be 40KN, 50MPa or 50N/mm^2 * 800mm^2 = 40,000N or 40KN, butI'm still stuck on the other link would it have the same thickness as link BD? link BD 800mm^2/510mm = 1.57mm? is it the same for the rest of the problem?
ok i redid the math in order for mpa to be converted to force I would have to convert all the measurements to meters so now I get 40N. but I am still stuck I'm not sure if that's the force or do I have to calculate the cross sectional area of ABC if so how do I go about doing that?
Knowing the portion of the link BD has a uniform cross-sectional area of 800mm^2, determine the magnitude of the load P for which the normal stress in that portion of BD is 50 Mpa. (Hint draw a free body diagram of the link ABC)
http://img137.imageshack.us/img137/9845/problmezf4.png
sorry for...
I was wondering if someone could check my work and see if its correct.
An object of mass 100.00kg moves in a straight line under the influence of a force given by F(t)=(120 N/s)t+4000N
At t=0 it is moving at 0 m/s. Determine its speed at t=5.00 s
What i did is this. I plugged 5s into the...
well I used the quotient rule, at the end because it seemed most logical. all we have to do is calculate the amount of joules whic m*N= J and we find the deriv. of the position time function given But, I'm not so sure.
the angle comes out to 5.71 which is almost the same, I tried that as well and if you add 1m to every 10m the answer is always the same .1 so I'm not sure how it would make a difference if used 1:10? so I'm still confused, also does anyone know if my formula is correct?
Hello again and here is another problem I can't seem to figure out.
Chelsea, a calico cat, runs in a straight line chasing a car down the street. Chelsea weighs 36.5 N. and her position along the road is given by
x(t)=(3.50m/s)t+(0.425m/s^2)t^2+1.00ms/(t+1.00s)
Calculate Chelsea's kinitec...
Hey hows everyone well I'm busting my brain over this problem,
On a level road the stopping distance for a certain car going 80km/hr is 32.0m. What would be the stopping distance for this car when going downhill on a 1:10 grade? A grade of 1:10 means the elevation drops 1.0m for a foward...