I'm assuming you at least have a graphing calculator. In function mode just do y1=250 and y2=100cos(arcsin(2sin(phi))) + 200cos(phi) = 250, which is what you had. Find the right range and domain and the intersection points will be your answer. you could also set it equal to 0 and fin the zeros.
well, this case does not have variables x or y in the limits. do the definite integrals separately and just multiple the results together. so it'll be like this c*\int_{0}^{w}xdx*\int_{0}^{h}dy
You have the right answer. To evaluate your expression, use dA = dxdy, so now you have \int\int{x*dxdy} with x going from 0 to w, and y going from 0 to h. If you go through the steps, you will get the same answer.
Sorry jarednjames, but that picture is correct. It's a copy out a picture given in the book. I went and talked to a physics professor today, and the way that I have it set up is correct. Sorry to waste all of your time.
Thanks for the replies. @Fizzynoob- that's the approach I took when I did it the first time, except I summed up forces for the block because the mass of the wedge is unknown.
I understand that the 8.4957 m/s/s is the acceleration of the block down the ramp. I don't believe that is the acceleration that I want, though. The question is asking for the acceleration of the wedge parallel to the ground. That's why I did the trigonometry at the end
So you're saying that since gravity is the only force pulling the block down, the acceleration component must be opposite that? I understand that. using that equation, you get an acceleration down the ramp of 8.495709211 m/s/s. Using trigonometry, a = ax/cos(60) where ax is 80495719211 m/s/s...
Homework Statement
A 2.0 kg block rests on a frictionless wedge that has a 60 deg incline and an acceleration \vec{a} to the right such that the mass remains stationary relative to the wedge. a) Draw the free-body diagram of the block and use it to determine the magnitude of the acceleration...