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ay negative pala assumption

Sir, bat po nag iba ng assumption nyo. sa first ang sabi nyo po 4Qb= tc/b + Yc. tas sa huli po nging Yc=4Qb + tc/b

ur probset are so far the best I've encountered sir, very thought provoking. same din sa hydraulics. nice problems

Good day engr. bakit nyo po gi consider na 0 ang support at right and hinge reactions while solving the Slope at C(left)? Thanks po

Thank you sir.

Thank u

you da man

This is really well done. Thank you

I dont anderstand thes . whet is thes ?

Yung problem 1 What is that 1mm/2 What is 2 in their sir?

thankyou so much sir!!!!

What's yur reference book?

What's your reference book?

Bakit po hindi nyo sinama yung 890sine theta(0.4178) at yung 63568.8...?

Thank you very much. You put this video up just twelve days before I needed it. That is lucky timing.

Which bookare u referring to?

Nice video Sir, pwede po ba malaman reference book niyo for this problem?

good noon po sir! mag ask lng ako about P-18, bakit po minimum si Force F when it has the same direction and angle with the Resultant of the two known forces?

Thank you for this sir!

Just wanted to take a moment to express my sincere gratitude for your dedication and support throughout this course. Your guidance has been invaluable, and I deeply appreciate all the effort you've put into our learning.

@elizaflores2260. What you got is 49.41 degrees not 49 deg. 41 minutes; press the degree key.

At 4:15 Corrections: P/(sqrt(5)-Q/sqrt(2)=-168.21, this is equation 2; Solving equations 1 and 2, P=86.76 lb, and Q=292.8 lb.

R1(6)=1.048&6)3+3.3(156/55), R1=4.704 KN ; Mmax=4.704*156/55)-1.048(156/55)(156/110) , Mmax=9.127 KN.m ; Mmax (w/ Impact)=9.127(1.3)=11.865 kN.m fb(w/ impact)=6(11/865)1000^/[150*300^2] fb(w/impact)=5.273 MPa < 8 MPa (beam is safe against bending) ; Vmax(6)=1.048(6)(3)+3.3(6-18/55) , Vmax=6.264 KN ; Vmax(w impact)=6.264(1.3) = 8.143 KN; fvmax*w pmpact)=[3*8143/(2^150^300)]=0.2714 MPa < 0.6000 MPa (beam is also safe against shear).

Good morning po sir, regarding problem 8. At around 24:15 I would like to clarify if there was a mistake regarding the MMax, because following the calculations in the video, the MMax should be about ~9.132 Kn*m. (11.872 Kn*m, if we add the Impact Factor). Thank you in advance po sir.

Good Evening sir, I would like to clarify something in Problem 8. Isn't the beam weight supposed to be 1.048 kN/m and not 1.408 kN/m? Thank you sir. 24:15

For Problem 3, the 3rd solution would be the application of the dot product: Fn=(1000i+7500j)dot(cos85i+sin85j)=7559 N and Ft=(1000i+7500j)dot(cos(-5)i+sin(-5)j)=342.5 N.

At 7:50, w=2(22.77*9.81)+2(77300)(0.375*0.01)= 1026 N/m. Please remove 2 in (2*0.375*0.01) so that it will only be (0.375*0.01). I already multiplied it by 2.

Thank you po engr/prof

Hello, sir. Would just like to clarify for the Moment of Inertia for the semi-circle. Should we not use 0.11r^4 po, or is that a different formula for the MI of a semi-circle po? Thank you so much for your response! 💕

in problem 2.1, why there is divided by 1000 in finding the pressure. i dont get it, because there is no unit

saw this on facebook! subscribed for future reference <3

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(b). NB'cos36.87-882.9-196.2=0; NB'=1348.88 N; MA=0=882.9x-1348.88(2)+196.2(2) ; x=2.611 m

In part (b) NB'cos36.87-882.9-196.2=0; NB'=1348.88 N; MA=0=882.9x-1348.88(2)+196.2(2) ; x=2.611 m.

Hello sir, good day, where could i watch the derivation vid for the formula po? Thank you

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Nice

Hello professor how did you know pressure at near end is zero ?

thank you sir

6:32 sir paano po nakuha ang 4000?

thanks for these contents sir

THANK YOU SIR MORE PWER TO YOU

Hello, engr! What if may additional given po like bending stress, pano po yung pagsolve? Thank you ❤

nice

Excellent!

thank you sir

what if the length 4x po is not given in the problem, what can we do to obtain the volume displaced? thank you po!

Thank you sa video mo sir, by the way sir any recommendations books in hydraulics

😢 'promo sm'

At 5:32, t=229.1 seconds not 221.1 s