Solución.
1.- Definir r y F
\(r1=4i+5j+3k\)
\(r2=4i+5j+3k\)
\(F1=100i-120j+75k\)
\(F2-200i+250j+100k\)
2.- Realizar producto cruz.
F1 X r2
= \(i\left(5\left(75\right)-120\left(3\right)\right)-j\left(4\left(75\right)-100\left(3\right)\right)+k\left(4\left(120\right)-100\left(5\right)\right)\)
=\(i\left(375+360\right)-j\left(300-300\right)+k\left(-480-500\right)\)
=\(755i-980k\)
F2 X r1
=\(i\left(5\left(100\right)-250\left(3\right)\right)-j\left(4\left(100\right)-200\left(3\right)\right)+k\left(4\left(250\right)-\left(-200\right)5\right)\)
=\(i\left(500-750\right)-k\left(400+600\right)+k\left(1000+1000\right)\)
=\(-250i-1000j+2000k\)
3.- Obtener resultado, ya que hay mas de dos momentos.
FT = \(\left(735i\ -250i\right)+\left(-1000j\right)+\left(-980k+2000k\right)\)
FT= \(485i-1000j+1020k\)
Problema 2
Two boys push on the gate as shown. If the boy at B exerts a force of FB= 30lb, determine the magnitude of the force FA the boy at A must exert in order to prevent the gate from turning. Neglect the thickness of the gate.