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3 years ago

En la figura, la tensión desarrollada en cada

Physics

1 answer:

malfutka [58]3 years ago

8 0

Answer:

Parte A

El ángulo con respecto al horizonte, de la segunda cuerda es de aproximadamente 19,47°

Parte B

La masa de la caja que se va a sostener es de aproximadamente 0,7808 kg.

Explanation:

Parte A

Los parámetros dados son;

La tensión en la cuerda, T₁ = 8 N

La tensión en la cuerda, T₂ = 6 N

El ángulo de inclinación de la primera cuerda con la horizontal, θ₁ = 45°

Sea θ₂ el ángulo de inclinación de la segunda cuerda, obtenemos;

T₁·cos (θ₁) = T₂·cos (θ₂)

∴ 8 N × cos (45°) = 6 N × cos (θ₂)

cos (θ₂) = 8 N × cos (45°) / (6 N) = (√2)/2 × (4/3) = (2·√2)/3

θ₂ = arcos ((2·√2) / 3) ≈ 19,47°

El ángulo con respecto al horizonte, de la segunda cuerda, θ₂ ≈ 19,47°

Parte B

El peso de la caja, W = T₁·sin (θ₁) + T₂·sin (θ₂)

∴ W = 8 N × sen (45 °) + 6 N × sen (19,47 °) ≈ 7,66 N

El peso de la caja que se va a sostener, W ≈ 7,66 N

La masa de la caja que se va a sostener, m ≈ 7,66 N / (9,81 m/s²) ≈ 0,7808 kg

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As Per Provided Information

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here

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