Answer:
2) τ = F x torque increases, 3) troque decreases,
4) man approaches the pivot and the child must move away
5) Σ τ = 0
Explanation:
2) when the man moves away from the pivot his torque increases significantly, since his distance increases
τ = F x
τ = m g x
therefore the system rotates faster
3) when the child approaches the pivot, his troque decreases, because the distance decreases
τ = f x
therefore the system must spin slower
4) If we place the man and the child on the side of a scale, the movement must be the man approaches the pivot and the child must move away, so that the torque is the same and the system can reach a balance
5) the rotational equilibrium expression
Σ τ = 0
is the one that describes the equilibrium of a system with several forces
Answer:
The potential energy of an object is the energy possessed by the object due to its position in a gravitational field.
Mathematically, it is given by:

where
m is the mass of the object
g is the acceleration due to gravity at the location of the object
h is the height of the object
As we can see, the potential energy depends on the height of the object, h.
However, this height is relative, since it can be measured using different reference levels.
Generally, the reference level is taken to be the ground; so for example, an object located at the top of a cliff 40 m above the ground, has
h = 40 m
But this value is of course relative, since we can also take the reference level to be the top of the cliff, so in that case, we would have
h = 0
Answer:
u = 1.77 m/s
Explanation:
Conservation of momentum
2250(10.3) + 3160u = (2250 + 3160)(5.32)
u = 1.77 m/s
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)