Determine the magnitude of F so that the resultant couple moment is 12 kN m

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Determine the magnitude of F so that the resultant
couple moment is 12 kN m, counterclockwise. Where on
the beam does the resultant couple moment act?

[@ompatel621]

error in the video: you forgot to use cos30(0.3)

[@sonikgain25]

Really helped thank you.

[@CeRz]

Correction: F = (12-8*1.2)/(0.3cos30)=9.24kN
He forgot the cos30. Also you don't need the reasoning for the x-components. The math will sort itself out if you simply find the perpendicular distance (or shortest distance) between the force F and the counterforce -F resulting in the same answer. It's good to understand why of course, but for calculations simply it is not necessary to argue for any components cancelling out because they're collinear, have opposite sense of direction and of the same magnitude.
Good luck!

[@mikeloto21]

question
F not suppose to be F cos 30?

[@chovy7630]

it should be F cos(30). he forgot it

[@muhammetklnc2908]

Greetings from Gebze Technical Un.

[@johnan98]

Problem 5.27 Consider the smooth uniform bar shown in (Figure 1) , which has a mass of 19 kg . Neglect the thickness of the bar. Determine the x and y components of reaction at A using scalar notation. Determine the magnitude of reaction at B.

[@markiechivange7711]

Not all heros wear a cap

[@samantharussell2439]

The x components cancel out, but so do the y components. That's the whole thing about couple moments. They're the same magnitude in opposite directions, so why can we cancel the x, but work with the y?

[@user-xn7sw2of1e]

Not sure if you're still interested, but the x-components of F cancel out because they act along the same line of action with the same magnitude and opposite direction, so the perpendicular distance between them is 0, and hence their couple moment is 0. Whereas the y-components are not on the same line of action and therefore do not cancel out. I struggled with this at first as well

[@michellerussell4349]

@@user-xn7sw2of1e thank you!

[@user-xn7sw2of1e]

@@michellerussell4349 no problem, good luck in your studies

[@harrisonvillareal9091]

@@user-xn7sw2of1e Amazing comment thank you!!!

[@LostDarkKingEU]

Fking hero!!! Tyx