Recent content by Frisky90

I have the following diff.eq. (d^2)y/d(x^2) + 2 dy/dx + y = 0 , but I don't have the answer, so could you please check. The general solution that I got is y= e^(-x)[Ax+B] If we have that x=0 when y=3 and dy/dx=1 , Then the particular solution is y= e^(-x)[4x+3]? Does someone know a good...

My question is related to PIC Microcontrollers. Q 1 I have to use port D of the PIC16F917A as an output port and write a simple program to control 8 LEDS to generate the binary-equivalent of the following decimal numbers: 119, 76, 14, 55 & 99 I know how to convert the numbers, but...

I don't understand how do you derive this expression for a? Could you explain? And if you plug in the numbers in it, you still don't get the correct answer.

I can't find a problem with the units.I've converted km/h to m/s. And I know that the textbook is correct :)

Homework Statement The brakes on a car are applied when traveling at 55km/h and it comes to rest uniformly in a distance of 50m. The mass of the car is 800 kg . Calculate the braking force and the time for the car to come to rest. Homework Equations F=ma v=u+at t=s/v=50*3.6/55=3.27s...

The centrifugal force on the ball is F=(mv^2)/r= 125 N The tension in the cables will vary depending on the position of the ball. It will go back and forth 180 degrees parabola ?? I don't know how to connect this to the moments equation.

It is a bit of both. We are just starting dynamics. I also noticed that it is said in the problem'' For the dynamic conditions described'' , so I guess this value should be used and the calculations you say should be made but I don't have any idea how.

Thank you for the prompt responses. In order to find the reactive pivot force (Q.d) I summ all the forces in the vertical -T1-T(W)- T1cos60+T(R)=0 -2452.5-1962-1226.25+T(R)=0 T(R)=5640.75 N Correct?

Ok, thanks for the advice. I changed the sign and used 1.2990 for z2. So, the new answer is 3303.9838 ?

Thanks for the FBD. I computed z2= 1.3m. M(T1) along the lifting cable 0.2*T1= 490.5 N 4782.375-T2*1.3+490.5=0 T2=4056.06 N?

New FBD: http://img200.imageshack.us/i/img5991sf.jpg/ M(T1)=2.452.2*1.45=3556.125 N moment along the center of the boom=(1/2*2.5)*1/2(Right Triangle rule)*1962=1226.25 N M(T2)=T2*1.5 3556.125+1226.25+T2*1.5=0 T2=-3188.25 ?? Pls tell me the exact way to do that if the answer is...

Ok,here is the FBD : http://img593.imageshack.us/i/img5989v.jpg/ Moment of the 2452.5 F = 1.45*2452.5=3556.125 N Moment of the 1962 F= 0.625*1962=1226.25 N Sum of moments= 4782.375 N = Ax (tension in support cable ? )

a) R2=1.25+0.2=1.45 ? c) \sum Fx=-4414.5+Fa(-->)=0 \sum Fy=-4414.5cos60+Fr=0

Thx, this clarified the things a bit. So, the distance between the projection of the center of the ball on the horizontal and pivot would be 1/2*2.5=1.25 m which is the answer to Q.a. Can you give me more info on how to sum the momments about A?

Thx,you are really helpful. Q.C : The angle about which the boom is inclined ( 60o in the example) doesn't make any difference for the tension in the support cable,so I am excluding it from the calculation? Then, F2= 200*9.81+250*9.81=4414.5 N ? I don't understand what are they asking...