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cdizzlesmith2511
24.11.2020 •
Physics
Con base en el siguiente ejercicio, responda a cada una de las siguientes preguntas:
Se aplica una fuerza cuya magnitud es de 120 N formando un ángulo de 30º con la horizontal sobre un bloque de 220N, como se ve en la figura. Si el bloque adquiere una aceleración cuya magnitud es de 2 m/s2.
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Ответ:
1) 104 N
2) 60 N
3) 45 N
4) 0.37
5) Falso
Explanation:
1) La componente de la fuerza en el eje x es 120 × cos (30 °) = 103,923 N ≈ 104 N
2) La componente de la fuerza en el eje y es 120 × sen (30 °) = 60 N
3) El valor de la fuerza resultante del eje horizontal = Masa del bloque × Aceleración del bloque
Peso del bloque = Masa del bloque × Aceleración debido a la gravedad
Masa del bloque = 220 N / 9,81 m / s² ≈ 22,426 kg
El valor de la fuerza resultante del eje horizontal = 22,426 kg × 2 m / s² ≈ 44,856 N ∴ ≈ 45 N
4) La fuerza de fricción = N × μ = La componente de la fuerza en el eje x - El valor de la fuerza resultante del eje horizontal
Dónde;
N = La fuerza que actúa normal (perpendicular) al suelo = El peso del bloque - El componente ascendente de la fuerza en el eje y = 220 N - 60 N = 160 N
μ = El coeficiente de fricción
La fuerza de fricción = N × μ = 104 N- 45 N = 59 N
∴ N × μ = 160 N × μ = 59 N
μ = 59 N / (160 N) = 0,36875 ≈ 0,37
El coeficiente de fricción = μ ≈ 0.37
5) falso; La fuerza normal y el peso no tienen el mismo módulo.
Ответ:
Explanation:
As B has a y component of -5, A must have a y component of 5 to have the result lie on the x axis.
As B has an x component of 3 and the result has an x component of -4, A must have an x component of -7
A = (-7, 5)
magnitude A = √(7² + 5²) = √74 ≈ 8.60
θ = arctan(y/x) = arctan(5/-7) ≈ 144.46° CCW from the positive x axis