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2 years ago

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23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n

orth with a magnitude of 130.0 N and Jennifer’s force is 32° east of north with a magnitude of 180.0 N. (a) Find the net force in component form. (b) Find the magnitude and direction of the net force. (c) If Andrea and Jennifer’s housemates, David and Stephanie, disagree with the move and want to prevent its relocation, with what combined force F → DS should they push so that the couch does not move?

1 answer:

PSYCHO15rus [73]2 years ago

4 0

Answer:

a) (95.4 i^ + 282.6 j^) N , b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

sin θ = Fjy / Fj

cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

T = 90 -32 = 58º

Fjy = Fj sin 58

Fjx = FJ cos 58

Fjx = 180 cos 58 = 95.4 N

Fjy = 180 sin 58 = 152.6 N

Andrea's force is

Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

Fx = Fjx

Fx = 95.4 N

Axis y

Fy = Fjy + Fa

Fy = 152.6 + 130

Fy = 282.6 N

F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

F² = Fx² + Fy²

F = √ (95.4² + 282.6²)

F = √ (88963)

F = 298.27 N

tan θ = Fy / Fx

θ = tan-1 (282.6 / 95.4)

θ = tan-1 (2,962)

θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

F' = 298.27 N

θ' = 180 + 71.3

θ = 251.4º

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Traslate

La velocidad en una onda en una cuerda es

v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

v= λ f

Usemos estas expresión en nuestro problema, para las condiciones iniciales

v= √ To/μ

√ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

T = 2T₀

√ ( T/μ) = λf

√ ) 2T₀/μ = λ f

√ 2 √ T₀/μ = λ f

substituimos

√2 ( λ₀ f₀) = λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

λ₀ = λ

f = f₀ √2

f = 435 √2

f = 615,2 Hz

b) La tension se reduce a la mitad

T = T₀/2

RA ( T₀/2μ) = λ f

Ra(T₀/μ) 1/ra 2 = λ f

fo /√ 2 = f

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<h3>Question:</h3>

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B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 $\sqrt{5.0^2 + 0.02^2}$ )

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