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Dominik [7]
3 years ago
8

Newton's laws of motion

Physics
2 answers:
Kipish [7]3 years ago
8 0

Answer:second law

Acceleration of an object depends on the mass of the object and the amount of force applied

Strike441 [17]3 years ago
4 0
It’s newton’s 2nd law of motion
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Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f
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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.

M_{o}=r_{A/O} * F_{AB} (producto cruz)

Necesitamos identificar los puntos:

O (0,0,0) [in]

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B (0, 24,8) [in]

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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

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]

De esta manera realizando el producto cruz tenemos

M_{O}=r_{A/O} * F_{AB}

M_{O}=0i-685.7j+2057.1k [Lb*in]

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

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]

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