Recent content by into space

Thanks for the reply. Hmm, I see what you are saying. So at s1=0, and the shear at point O is zero, then qs1(0)=-3*Vz*t*(0)^2/(10*Iyy)+q(0)=0 which means q(0) has to equal 0. Can you explain how you know there has to be an open section just from looking at the shear center?

Homework Statement Homework Equations qs1(0)=qs3(5) qs2(0)=qs1(5) qs3(0)=qs2(6) Vy=0 q(s)=-Vy/Iyy*Q(s) Where Q(s)=y*A Where y=y coordinate of centroid and A=Area The Attempt at a Solution To be honest I'm not sure how to separate the q(0) term, nor do I know what the...

Awesome, I got the right answer! Thanks so much, I had completely forgot eccentric anomaly was solved differently depending on the orbit/trajectory.

So I'll post a solution I've been working on: I'm not sure how to think of flight path, but I assumed the transfer maneuver was like a Hohmann Transfer with 60 DU forming the apoapsis radius: e = (60-1.25)/(60 + 1.25) = .9592 Now I'm trying to find the semi-major axis from the periapsis...

Homework Statement A lunar probe is given just escape speed at an altitude of .25 DU and flight path angle 30°. How long will it take the probe to get to the vicinity of the moon (r = 60 DU), disregarding the moon's gravity? Circular orbit, radius = 1.25 DU θ = 30° gravitational parameter =...

So the initial phase shift is -.466 rad and the amplitude is .0089 m. In order to get the correct initial phase shift from the book I have to add -.466 and 3.14 (pi) which yields 2.67 rads (this is the answer given in the book). Why do I have to add my answer in order to get the desired answer...

Homework Statement A wave travels along a string in the positive x-direction at 30 m/s. The frequency of the wave is 50 Hz. At x = 0 and t = 0, the wave velocity is 2.5 m/s and the vertical displacement is y = 4 mm. Write the function y(x,t) for the wave. Homework Equations velocity =...

And so the monster of the problem is defeated. Again, thanks for all the help you two. If this problem comes up on a test, I'll blast it out of the water. :)

Haha you're right, I kind of got ahead of myself in anticipation of solving a problem I've been working on for more than four hours. And yes, thank you ehild for the help as well. Though I know now how to solve this problem using double integrals, I really wanted to know how to solve it with...

Makes perfect sense now. Thank you. And the 1/2 was on the bottom so I flipped it up to the top as a 2 then brought the 1/3 out and it became 2/3.

Huh. The way I'm thinking of it is like this. The area of the large sector is shaded in blue in this picture. The area of the small sector is shaded in yellow here. And the area of he combined sectors is overlapped in green here. Wouldn't it be redundant to include the smaller...

Ok, dividing that by the total area which is: A = .5(pi*(r2)^2 - pi*(r1)^2 - 2alpha(r2)^2) And recognizing from trig identities that cos (pi - alpha) = -cos(alpha) the equation is now: [(r2)^3 - (r1)^3][2cos(alpha)]/[pi*(r2)^2 - pi*(r1)^2 - 2alpha(r2)^2] My answer differs from the...

Ok, just to make sure I'm doing this right, the end result of the double integral of sin(θ)*(r^2) dθ dr from θ = alpha to θ = (pi - alpha) and r = r1 to r=r2 is: [cos(alpha)*r^{3}_{2}*(1/3) - cos (pi - alpha)*r^{3}_{2}*(1/3)] - [cos(alpha)*r^{3}_{1}*(1/3) - cos (pi - alpha)r^{3}_{1}*(1/3)]

The problem I have is the formula for a circular sector has the middle of the sector resting on the x-axis, with both angles (alpha) equally above and below the x-axis. This makes the y-bar for the sector centroid 0. Here's a picture hopefully describing it better: Here There are two sectors...

I'm unsure of how to go about solving an integral when there is a dr and a dθ. Are you using double integrals? Because we haven't learned that yet.