Tarzan, who weighs 771 N, swings from a cliff at the end of a 18.2 m vine that h

Question

Tarzan, who weighs 771 N, swings from a cliff at the end of a 18.2 m vine that hangs from a high tree limb and initially makes an angle of 24.1° 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 704 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?

This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3gm/s^3times

Explanation / Answer

Given that,

Weight of Tarzan = W = 771 N ; so his mass will be = m = W/g = 771/9.8 = 78.67 kg ;

length of vine = l = 18.2 m ; angle = theta = 24.1 Deg ; Tension = T = 704 N ;

(a)The force from vine on Tarzan is given by:

F = Tx + Ty

F = T sin(theta) i + T cos(theta) j

F = 704 x sin(24.1) i + 704 cos(24.1) j

F = ( 287.46 i + 642.64 j ) N

(b) F(net) =( Fnet x )i + (Fnet y) y

Fnet - x = Tx = 704 x sin(24.1) = 287.46

Fnet -y = Ty - W = 704 - 771 = -67 N

F(net) = 287.46 i - 67 j

(c) the magnitude will be given by:

lFl = sqrt [ (287.46)2 + (-67)2 ] = 263.71 N

Hence, magnitude = F = 295.16 N

(d) Direction will be given by:

alpha = tan -1 ( -67/287.46) = - 13.12 Deg = 360 - 13.12 = 346.88 deg

Hence, direction = alpha = 346.88 Deg

(e) We have F = 295.16 N

We know that, F = ma => a = F/m = 295.16/78.67 = 3.75 m/s2.

Hence, a = 3.75 m/s^2.

(f) The direction acceleration will be same as that of net force.

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