Answer:
One horse pulls a ring to the north with a force of 2000 N, and another to the east with an F of 3000 N. With which F has to pull a third horse and where to so that the ring is in balance.
Step-by-step explanation:
Given that, a horse pulls a ring to the north with a force of 2000N
F_1 = 2000 •j
Another horse pull in the east ward direction with a force of 3000N
F_2 = 3000 •i
A third horse that balanced the system force and direction.
F_3 = ?
Using equilibrium principle, the sum of acting on the ring is 0N
Then,
F_1 + F_2 + F_3 = 0
F_3 = -F_1 - F_2
F_3 = -2000•j - 3000•i
F_3 = -3000•i - 2000•j
Then,
The magnitude of the force is
|F_3| = √(-2000)² + (-3000)²
|F_3| = √(4,000,000 + 9,000,000)
|F_3| = √13,0000,000
|F_3| = 3605.55 N,
That is the magnitude of the third horse
Then, it's direction can be calculated using
tan θ = (y / x)
θ = arctan(y/x)
θ = arctan(-2000/-3000)
θ = 33.69°
But this is in the third quadrant because the both direction of x and y is negative.
Then, the direction is W33.69°S
Or total angle is 180 + θ = 180 + 33.69
Direction = 213.69°
In Spanish
Dado eso, un caballo tira de un anillo hacia el norte con una fuerza de 2000N
F_1 = 2000 • j
Otro tirón de caballos en dirección al barrio este con una fuerza de 3000N
F_2 = 3000 • i
Un tercer caballo que equilibraba la fuerza y dirección del sistema.
F_3 =?
Usando el principio de equilibrio, la suma de actuar en el anillo es 0N
Entonces,
F_1 + F_2 + F_3 = 0
F_3 = -F_1 - F_2
F_3 = -2000 • j - 3000 • i
F_3 = -3000 • i - 2000 • j
Entonces,
La magnitud de la fuerza es
| F_3 | = √ (-2000) ² + (-3000) ²
| F_3 | = √ (4,000,000 + 9,000,000)
| F_3 | = √13,0000,000
| F_3 | = 3605.55 N,
Esa es la magnitud del tercer caballo.
Entonces, su dirección se puede calcular usando
tan θ = (y / x)
θ = arctan (y / x)
θ = arctan (-2000 / -3000)
θ = 33.69 °
Pero esto está en el tercer cuadrante porque ambas direcciones de x e y son negativas.
Entonces, la dirección es W33.69 ° S
O el ángulo total es 180 + θ = 180 + 33.69
Dirección = 213.69 °