
#1
Aug2107, 10:17 PM

P: 38

http://wps.prenhall.com/wps/media/ob.../9e_16_43.html
That is the image of the problem, which includes a solution. PROBLEM Besides being really confused on their work, the solution they give and the solution in the back of my book are both different! They said Wcd is 15.1 rad/s and my book says 4.03 rad/s, they have the same dimensions and everything. heres what im doing, despire the crazy solution given.. Im just trying to get some sort of method and understanding of this stuff, Im taking this course in 5 weeks so its sort of a rush to get it all in.. I want to find Angular velocity of link CD at the instant shown Vc=Vd+Vc/d ; Vc=Vd+(Wcd x Rcd) Vd wont be moving so it goes to zero So Vc=(WcdxRcd) ((((So I must need to solve for Vc to get Wcd)))) Va= Vb+ Va/b; Va= Vb + (Wab x Rab), Va wont be moving so it goes to zero Vb= (Wabx Rab), I can solve for this because I am given Wab and Rab, Vb=18 in/s Vc=Vb+Vc/d; Vc=Vb+(WcdxRcd) Vc= 8 + (WcdxRcd) So I just need to find Vc somehow.. On the solution given they do some insane trig that I just dont get, your input is much appreciated so thanks ahead.. 



#2
Aug2207, 07:47 AM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,388

I agree with your book's solution.
First suppose D is free to move, and write down the motion of points B C and D in terms of the angular velocities (positive anticlockwise) Vb = (0, 18) in/sec Vc = Vb + (8Wbc cos 30, 8 Wbc sin 30) Vd = Vc + (4Wcd sin 45, 4 Wcd cos 45) But Vd = 0 since it is pinned. So 8 Wbc cos 30 + 4 Wcd sin 45 = 0 18  8 Wbc sin 30 + 4 Wcd cos 45 = 0 Eliminating Wcd (remembering sin 45 = cos 45) gives 18  8 Wbc (sin 30 + cos 30) = 0 Wbc = 1.647 rad/s And Wcd = 2 Wbc cos 30/sin 45 = 4.034 rad/s I didn't try to find what why this is different from the web page solution. 


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