<p>Tarzan, who weighs 854 N, swings from a cliff at the end of a 18.9 m vine tha

Question

<p>Tarzan, who weighs 854 N, swings from a cliff at the end of a 18.9 m vine that hangs from a high tree limb and initially makes an angle of 21.1&#176; 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 797 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?</p>

Explanation / Answer

Weight of the tarzan is W = 854 N Lenth of the vine is 18.9 m Tension in vine with out tarzan is T   = 797 N a ) Force from the vine   T = T cos ( 90 - 21.1 ) i + T sin ( 90 - 21.1)j                                            = 797 cos 68.9 i + 797 N sin 68.9 j                                            = (286.9 i + 743.5 j )N b ) Net force acting on tarzan is   F_net = T + W                                                  =(286.9 i + 743.5 j )N - 854 j                                                  = ( 286.9 i - 110.4j) N c ) Magnitude F_net = (286.9)^2 + ( -110.4)^2                                      = 307.4 N d ) direction = tan^-1 ( -110.4 / 286.9)                          = -21.0below horizontal e ) Magnitude of acceleration   is F = ma                                       a = F_net / w / g                                           = 307.4/ 87.1                                           = 3.52 m/s^2 f ) direction of tarzan's acceleration is parallel to direction of force hence                 21.0 below horizontal   Weight of the tarzan is W = 854 N Lenth of the vine is 18.9 m Tension in vine with out tarzan is T   = 797 N a ) Force from the vine   T = T cos ( 90 - 21.1 ) i + T sin ( 90 - 21.1)j                                            = 797 cos 68.9 i + 797 N sin 68.9 j                                            = (286.9 i + 743.5 j )N b ) Net force acting on tarzan is   F_net = T + W                                                  =(286.9 i + 743.5 j )N - 854 j                                                  = ( 286.9 i - 110.4j) N c ) Magnitude F_net = (286.9)^2 + ( -110.4)^2                                      = 307.4 N d ) direction = tan^-1 ( -110.4 / 286.9)                          = -21.0below horizontal e ) Magnitude of acceleration   is F = ma                                       a = F_net / w / g                                           = 307.4/ 87.1                                           = 3.52 m/s^2 f ) direction of tarzan's acceleration is parallel to direction of force hence                 21.0 below horizontal  

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