Question

I got vc = 3.646 and w = 12.93. w is magnitude of angular velocity. This was correct. I got ac = -134.31 and a(magnitude of angular acc) = -277.37. This was correct. I’m having trouble finding the acceleration of point A. i got the x part right but not the y part. I got aA = 127.75i + 43.8j but the 43.8j part is wrong.

hapter 5, Problem 5/108 Multistep has a downward velocity v-3.0h/sec and an upward acceleration a " 2.2 ft/see when the d we s "the poston shown, magnitude of the acceleration of point A. Roller C is in continuous contact with the inclined surface erm re th 405 5" 18

Answer 1

Ft /sec 2. Now, Rol les moves along the inclined C. vet -ve sin 1 8 Hri te Relative velocity Equa-tion-f0γ Points B and c

Position vector, Bc = +/5co54ox 12 ft sin 40 12 Eq. ( be comes Equating components -0.268 Eq CA) Fquatng comPonents Ve sin 18- 31guACq(3)

Solve Eqs (A) and (B) V3.64S FtIse c 126 Ac Angular velocit oF Yod AC sec NOw to find aA, Wite Relative C2) Now, we have to Find the angular accejetAB

Relative Accele Equation or B 4 c rG) 2. AC 2 - (12 935) (o 31g -o 203j ナ0.268 KAB人 ー53343人 ナ44.840」 Equating components Equating J tomPonents

solve Eq s CC) 4 (d) n(AB -273-602 Yad/Sec2. = + .5 3.34 44.840 J. 2 se c sec