Question

I need help with the algebra if the problem. I saw solutions showing the correct final answer but I do not get the final answer they get. Please show the algebra of the last third of the problem. So I understand what do I do wrong.

Answer: 0.587 rad/s^2

1. The 50-kg uniform crate rests on the platform for which the coefficient of static friction is us = 0.5. If the supporting links have an angular velocity w = 1 rad/s, determine the greatest angular acceleration a they can have so that the crate does not slip or tip at the instant e = 30°. m= 1 rad/s 4 m / //4 m LA 1.5 m 0.5 m

Answer 1

Mass of crate= 50 kg. Coefficient of static flichen, = 0.5 . Tangential acceleration, 9+ = 910. Normal acceleration, an= 99 w2 an 130° Boyan B. q. - - - - 13 al Sak 9x = 14 cos 30° + 91 02 sin 30° = 4x2 3 + 4x0² x Į = 3.4644. + 2 0 Qy = 9602 00830°- 4* Sun 30° = 4X07² B - 4. 1 - 3.464-22 -@ Case(1) when she crate is on the verge of slipping Fmax = MgN=0-5N +1 Efy = m (anly. N-509 - 50 ay. using equation © Foment in N-509 = 50 (3.484-22) N = 50g+ 50 (3.464-20) =490-5+173-2-1000 = 6637-1004 - ③ moly max Efx = m (ode: Fmax = 509x. 0.5N = 50 (3.4642+2) Meing 020 0-5 (6637-1004) - 50( 3.4642+2) 331.85-502 = 173-24 loo

0.5m. man. LW 1 sm 331.85-100 = 173.22 +50d. 231.85 = 223-22 2 = 1,0387 rad /s2 Case (2) when Gate is on the verge of sipping EMP = (EMKJP 50g (0.25) - Sour (). So ay (0-25) 50x9:31 X0.25 = 50 (3.4644+2) (0-75) - 50(3.464-20) (-25) 2012:62512990V * 45-673-09-2004 mar T 9,81X0.25=205980+ 1.5-0.866+0.5%. = 3.0980+ 0.634 1 x = 0.587 sead/5² from the above two values, the maximum angulee acceleration that will forevent both sipping and slipping =0-587 stad/s2) W