Tarzan, who weighs 819 N, swings from a cliff at the end of a 22.5 m vine that h

Question

Tarzan, who weighs 819 N, swings from a cliff at the end of a 22.5 m vine that hangs from a high tree limb and initially makes an angle of 25.1° with the vertical. Assume that an x axis points horizontally away from the cliff edge and ay axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is 742 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?

Explanation / Answer

25.1 degree from vertical , 64.9 from horizontal

T = 742 N

T = Tcos 64.9 + T sin 64.9 = 324.1i + 667.47j

part b )

No longer in contact with the cliff, the only other force on Tarzan is due to earth’s gravity (his weight).

Fnet = T + W = 324.1i + 667.47j - 819j

Fnet = 324.1i - 151.53j N

part c )

|F| = sqrtx^2+y^2)

|F| = 357.77 degree

part d )

theta = tan^-1(y/x)

theta = -25.058 degree = 334.94 degree CCW

part e )

a = F/m

m = W/g

a = 4.28 m/s

part f )

acceleration in force direction

theta = 334.94 degree

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