Current Density is the amount of current that a conductor can safely carry without undue heating per cross - sectional area.
Current Density = Current carrying capacity/Area
Answer:
Moc = -613.25 [lb*in]
Explanation:
Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.
Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.
(producto cruz)
Necesitamos identificar los puntos:
O (0,0,0) [in]
A (12,0,0) [in]
B (0, 24,8) [in]
C (12,24,0) [in]
![r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k](https://tex.z-dn.net/?f=r_%7BA%2FO%7D%3D%2812%2C0%2C0%29%20-%20%280%2C0%2C0%29%5C%5Cr_%7BA%2FO%7D%20%3D%2012%20i%20%2B%200j%2B0k%20%5Bin%5D%5C%5CAB%20%3D%20%280%2C24%2C8%29%20-%20%2812%2C0%2C0%29%5C%5CAB%20%3D%20-12i%2B24j%2B8k%20%5Bin%5D%5C%5C%5BLAB%5D%3D%5Cfrac%7B-12i%2B24j%2B8k%7D%7B%5Csqrt%7B%2812%29%5E%7B2%7D%20%2B%2824%29%5E%7B2%7D%20%2B%288%29%5E%7B2%7D%20%7D%20%7D%5C%5C%20LAB%3D-%5Cfrac%7B3%7D%7B7%7D%20i%2B%5Cfrac%7B6%7D%7B7%7Dj%2B%5Cfrac%7B2%7D%7B7%7Dk)
El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].
![F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]](https://tex.z-dn.net/?f=F_%7BAB%7D%3D-%5Cfrac%7B600%7D%7B7%7D%20i%20%2B%5Cfrac%7B1200%7D%7B7%7Dj%2B%5Cfrac%7B400%7D%7B7%7D%20k%20%5BLb%5D)
De esta manera realizando el producto cruz tenemos

![M_{O}=0i-685.7j+2057.1k [Lb*in]](https://tex.z-dn.net/?f=M_%7BO%7D%3D0i-685.7j%2B2057.1k%20%5BLb%2Ain%5D)
Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.
![OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j +0k](https://tex.z-dn.net/?f=OC%20%3D%20%2812%2C24%2C0%29-%280%2C0%2C0%29%5C%5COC%3D%2012i%2B24j%2B0k%5BLb%5D%5C%5CLOC%20%3D%20%5Cfrac%7B12i%2B24j%2B0k%7D%7B%5Csqrt%7B%2812%29%5E%7B2%7D%20%2B%2824%29%5E%7B2%7D%20%2B%280%29%5E%7B2%7D%20%7D%20%7D%20%5C%5CLOC%3D%5Cfrac%7B12%7D%7B%5Csqrt%7B720%7D%7Di%2B%5Cfrac%7B24%7D%7B%5Csqrt%7B720%7D%7Dj%20%20%2B0k)
El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC
![M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]](https://tex.z-dn.net/?f=M_%7BOC%7D%3DL_%7BOC%7D%2AM_%7BO%7D%5C%5CM_%7BOC%7D%3D%28%5Cfrac%7B12%7D%7B%5Csqrt%7B720%7D%7Di%20%2B%5Cfrac%7B24%7D%7B%5Csqrt%7B720%7D%7D%20j%2B0k%20%29%2A%20%280i-685.7j%2B2057.1k%29%5C%5CM_%7BOC%7D%3D%20-613.32%5BLb%2Ain%5D)
Answer:
TRUE
Explanation:
The answer is true.
Balance forces acting on a body will not change the motion of the body because the body experiences no net resultant force in one direction. When any body experiences equal forces with opposite directions, the net force or the resultant force experience by the body is zero.
In case of an unbalanced forces, there is a net force acting in one direction and so it causes the body to change in its state of motion in the direction of the net force.