Tarzan, who weighs 852 N, swings from a cliff at the end of a 15.8 m vine that h

Question

Tarzan, who weighs 852 N, swings from a cliff at the end of a 15.8 m vine that hangs from a high tree limb and initially makes an angle of 20.0° 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 801 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

a)

rope is making angle = 20 with the vertical

thus we conclude that angle = 80 with the horizontal

F=(Tcos(80))i + (Tsin(80))j

T = 801

F = (139.092 N) i + (788.831 N) j

b)

Weight of Tarzan W = (852 N)(-j)

Fnet = F + W = (139.092 N) i + (788.831 N) j + (852 N)(-j)

Fnet = (139.092 N) i + (63.1689 N) (-j)

c)

magnitude of the net force

Fnet = root((139.092)^2 + (63.1689)^2)

Fnet =root (23336.8943)

Fnet = 152.7641 N

d)

direction of the force

tan(theta) = -(63.1689) / (139.092) = -0.45415192

theta = - 24.4252

e)

mass of tarzan is m = 852 / (9.8) = 86.938

acceleration of tarzan a = Fnet / m = 152.7641 / 86.938 =1.75716 m/s^2

f)

direction of net acceleration is same as the direction of net force

So,tha angle will be 24.4252 below the horizontal

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