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Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes an angle of 26.9° with the vertical. Assume that an x axis points horizontally away from the cliff edge and a y axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is 757 N. Just then, what are (a) the force from the vine on Tarzan in unit-vector notation, and (b) the net force acting on Tarzan in unit-vector notation? What are (c) the magnitude and (d) the direction (measured counterclockwise from the positive x-axis) of the net force acting on Tarzan? What are (e) the magnitude and (f) the direction of Tarzan's acceleration just then?

Respuesta :

Answer:

Part a)

[tex]T = 342.5 \hat i + 675\hat j[/tex]

Part b)

[tex]F_{net} = 342.5\hat i - 174\hat j[/tex]

Part c)

[tex]F = 384.2 N[/tex]

Part d)

[tex]\theta = 333 degree[/tex]

Part e)

[tex]a = 4.4 m/s^2[/tex]

Part f)

[tex]\theta = 333 degree[/tex]

Explanation:

Part a)

Magnitude of tension force is given as

[tex]T = 757 N[/tex] at 26.9 degree with vertical

[tex]T = 757 sin26.9 \hat i + 757 cos26.9 \hat j[/tex]

[tex]T = 342.5 \hat i + 675\hat j[/tex]

Part b)

Net force on Tarzen is given as

[tex]F_{net} = T + F_g[/tex]

[tex]F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j[/tex]

[tex]F_{net} = 342.5\hat i - 174\hat j[/tex]

Part c)

magnitude of the force is given as

[tex]F = \sqrt{F_x^2 + F_y^2}[/tex]

[tex]F = \sqrt{342.5^2 + 174^2}[/tex]

[tex]F = 384.2 N[/tex]

Part d)

Direction of the force is given as

[tex]tan\theta = \frac{F_y}{F_x}[/tex]

[tex]tan\theta = \frac{-174}{342.5}[/tex]

[tex]\theta = 333 degree[/tex]

Part e)

Magnitude of the acceleration

[tex]a = \frac{F}{m}[/tex]

[tex]m = \frac{849}{9.81} = 86.5 kg[/tex]

tex]a = \frac{384.2}{86.5}[/tex]

[tex]a = 4.4 m/s^2[/tex]

Part f)

Direction of acceleration is same as the direction of the force

[tex]\theta = 333 degree[/tex]

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