Question

At the instant shown, the length of the boom AB is being increased at the constant rate of 0.2 m/s, the boom is being lowered at the constant rate of 0.08 rad/s, and the truck is moving forward with a speed of 0.3 m/s and is slowing down at 0.05 m/s2. Determine (a) the velocity of point B, (b) the acceleration of point B

6 m O = 30° Fig. P15.164 and P15.165 15.165 At the instant shown, the length of the boom AB is being increased at the constant rate of 0.2 m/s, the boom is being lowered at the constant rate of 0.08 rad/s, and the truck is moving forward with a speed of 0.3 m/s and is slowing down at 0.05 m/s². Determine (a) the velocity of point B, (b) the acceleration of point B.

The answer on the back of the book is 15.165 (a) 0.603 m/s ⦪ 31.6°. (b) 82.0 mm/s2 ⦫ 34.9°.

Answer 1

2 ell V-6m Sap - ? 60. 0.3mys w=0.08 rad, 30.) B at-o-osm/ Acceleration of a reative to Boom = o abl boom = Coriolis accelaration 2 wy=2x8.08 X2=0.032 m/s² acceloration conciding to boom = aß= rot- 6x0.082 a's = 0.0384 ways? acclaration buom worito goound aBoom = -o'os m/ge so aceleration point B 30° apz = 0.032 60560° -0.05 -0.0384cm. 0.05m/s2 ABU = – 0.06725 miz A = -0.0384 sin 60'- 0.032 sy'n 38 ay 9 = -0.04691 m2.cz 0.0384 m/? AB Í - 006725)?+€0.0469172 0032 aRa AB= 0·082 m/s2, ap= 82 mm / 32 -0.0469) tomp = 0.0 6725 | B= 34. go

Nole. :- Unfrostmately I did not get some senswer as yours for relocity of poinB, But & get correct answer for for acclaration velocity coneinding to Boom with respect to breek = rw VB'- 6x0:08 - 0:48 m/s Vook. VR point B relative to boom wor.t. truck 8.2 m/ NB/boom- 0.2 m/s 30 0.2 m/s cos 30° Vasa = 0.3+0.48 cos 60+ o 2 VBA - 0.7132 m/s 430 0.3 m/s 0.48 sin 60 – 0:2 sin 300 VBy = 0: 3157 m/s 30.48 mis velocity of VB = Vusa +ve 10.7132? +0.31572 VB = 8-78 m/s 0.3157 temp = 0.7132 0.7132 2 ß 0.3837 B= 23.87°