Recent content by dch

Thanks grey_earl, so conclusion: 100 chips are manufactured in the beginning. Of these, 92 are good and 8 are bad. 99.75% of the 92 good chips pass the test = 91.77 good chips. 95.12% of the 8 bad chips pass the test = 7.61 bad chips. Altogether, 91.77 + 7.61 = 99.38 chips pass the test. The...

Homework Statement A manufacturing process produces 92% good chips (G) and 8% bad chips (B). The lifetime, in seconds, of chips is exponentially distributed E(\lambda). For good chips, \lambda1=1/20000 For bad chips,\lambda2=1/1000 Every chip is tested for 50 seconds prior to leaving the...

In Prob2, can I divide the shape to 2 rectangular : The centroid of upper rectangular is located at (0,9/2) & the other one is (0,2a) -> The centroid of the shape is located at (0,43a/14) right. How can I find the moment of inertia for the area about the x-axis.

So you mean: Ex+P2=0 Ey+Fy+FBD=0 where Ey=Fy

If there's a vertical force at F (eg Fx) it will be : Ex+P2-Fx=0 Ey+FBD=0 I think I can't solve this.

Homework Statement Please help me do these : [PLAIN]http://broadcastsol.com/321.JPG Prob1 The mechanical system of massless members is shown in the figure. Determine the relationship between two forces P1 and P2 to keep the system in equilibrium at position as showing in the...

Thank you so much! Now I have to do 2 problems and I don't want to make a new topic (very sorry about that because I thought I found my kindhearted teacher ^^ ). Please help me do these : [PLAIN]http:///broadcastsol.com/321.JPG Prob1 The mechanical system of massless members is...

M(D) = -P*DC*sinα-FBEcosα*DE*cosα-FBEsinα*DE*sinα = 0 -> FBE=P*DC*sinα/DE=3Psinα/2 D: Dx-FBE*cosα=0 Dy-FBE*sinα-P=0 -> Dx,Dy A: Ax-F(BE)cosα=0 Ay-F(BE)sinα - q =0 M(A) = 0 = -q*AB/2 + F(BE)sinα * AB =0 -> q/2 = FBE sinα -> Ay = 3q/2Is this right :D Thanks.

[PLAIN]http://broadcastsol.com/123.JPG Here's my solution for prob2: M(D) = -P*DC*sinα-F(BE)cosα*DE*sinα-F(BE)sinα*DE*cosα = 0 -> F(BE) D: Dx-F(BE)*cosα=0 Dy-F(BE)*sinα-P=0 -> Dx,Dy Q= Integral (q*da) at midpoint of AB A: Ax-F(BE)cosα=0 Ay-F(BE)sinα=0 ->Ax,Ay...

so that CD is equilibrium thus, F(DG)cos(45)-F(EF)=Cx F(DG)sin(45)=Cy Is this right :D In problem 2, the load q is just like the weight of beam AB or not. Thanks!

Can you give me some hints? :D I just thought that : -P+F(DG)cos(45)+Bx+Cx=0 -F(DG)sin(45)+By+Cy=0 Thanks

F(FE)xa=F(DG)xcos(45)x(a+b) right? Thanks

sorry How can I do this with the angle 45 at DG & DC thanks

Prob 1, i just found the force EF = P*(a+c)/a and stuck at C & DG Prob 2, I could not understand what "q" means. Thanks for your reply :)

Homework Statement [PLAIN]http://broadcastsol.com/123.JPG Prob. 1. The structure of massless members is shown in the figure. Determine the reaction forces at the supports C, and forces in the rod EF and DG. Givens: P, a, b, c. Prob. 2. The structure of massless members is shown...