Dynamics of Rigid Bodies - Rectilinear Translation | Engineering Mechanics |

[@joshycasuga2555]

Lupet ng content nyo sir assorted tlgang pang Masang Engineering topics galing!!

[@AvatarAang-bf1sj]

Thank you Sir. Magtatake na ako ng dynamics. This is very helpful po.

[@ronnmanalo9387]

sir tanong lang po dun sa 2nd problem. Same given po sa lecture namin kaso po sa solving 5s ginamit ng prof ko po para makuha yung velocity. kasi daw po in 10s yun na po yung time paakyat at pababa ng bato. ano po ba mas tama?

[@ericaarrubioortiz6276]

Nagkamali lang si sir pero what he really mean is that ung sa no.2 t=5sec making Vi=49.05m/s

[@easymoney7344]

Sir Baka may video Ka po (Dynamics Inclined ODE )❤️❤️ exam po namin next week Ito nalang po talaga Di ko pa na Aral❤️. Salamat talaga sir ❤️ na perfect ko po yung long quiz namin sa DE ❤️

[@engr9118]

Hi, sir! Sa sample no. 2 niyo po, bakit po yung pagbagsak na ng stone yung naging reference noong height(distance) na yung kinukuha? Bakit po hindi yung initial velocity na nakuha yung ginamit? Tyia po. Palagi po ako nanonood ng lec vids niyo para magreview lalo pag hindi ko magets yung ibang parts ng libro ni Singer 😁

[@kingharounsultan9220]

Tanong lang po sana sa example #2. If we set the final velocity at the top na 0, then the stone will have two separate times, on the way up and on the way down since at 10 seconds, it's already on the ground. If we calculate it that way, using t=10 seconds, di po ba ibig sabihin nyan eh nasa taas pa yung stone at t =10 seconds?

[@aervinapiag1785]

(2)

[@psychopath8385]

@@aervinapiag1785 Vo= 49.05 m/s and S=122.625m din sagot nyo dyan?

[@jacobyrrjebagongon7454]

Actually yung 10s kasi kasali na dun yung going up bali 5s and going downward is 5s

[@danielagbanlogjr9140]

Yong number 2 po ata is 5seconds pataas 5 seconds pababa

[@andressheenamaeb.7007]

more solving pa po engr about dynamics

[@sayonaraxvi668]

Problem #3 Ball 1 free fall s=Vot+1/2gt^2 Yf-Yo=Vo+1/2gt^2 (80-h)-80=1/2(-32)t^2 -h=-16t^2 h=16t^2 ----------->eqn 1 Ball 2 upward velocity s=Vot+1/2gt^2 Yf-Yo=Vo+1/2gt^2 (80-h)-0=40t+1/2(-32)t^2 -h=40t-16t^2-80 h=16t^2-40t+80----------eqn2 Sub eqn 1 to eqn2 16t^2=16t^2-40t+80 t=2s s(t)=16t^2 s(2s)=16(2)^2 s=64ft ---------answer The height relative to the ground since ground is set to s=0, and 80ft is measured from the ground.

[@sayonaraxvi668]

Must also be noted that even after 1 second, the ball 2 must have traveled higher than 40ft

[@banquilayjohnisaacb.1544]

sir sa problem no.3 Bakit po negative ung Gravity ng Ball 1 ? hindi po ba Positive dapat kasi Free falling body po sya meaning along the gravity po sya not against the gravity?

[@esmailbalogan3943]

Sir sa number 2 problem same lng ng sagot if yung gagamitin na i iniatial velocity is 98.1 then times sa 10. Same lng ng sagot na lumalabas pag yung value ng Vo is ganun po sa ginawa nyo which is 0.

[@esmailbalogan3943]

Ask lng din po if tama rin na gamitin yung 98.1 as iniatial velocity

[@frichjunvisto1931]

sir ano po reference na book nyo for this ?

[@nellelingacion1711]

231.649m po sagot sa no.5 engr

[@KiriGiri-fd7zq]

ano po ba tama sa no. 2?

[@kopikazuya]

Sir ano po gamit niyo po na book?

[@abatandchill]

Engineering Mechanics by Singer

[@engr9118]

yung blue po?

[@abatandchill]

@@engr9118 Yes :)