Question

PLEASE HELP

HISSAGE MY INSTRUCTOR IULİ SCREEN PRINIERVERSION 4BACK NEXT Tarzan, who weighs 820 N, swings from a cliff at the end of a 19.1 m vine that hangs from a high tree limb and initially makes an angle of 23.9° 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 750 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? (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number Units j Units Units Units Units Units Activate Windows Go to Settings to act LINK TO TEXT

Answer 1

(a)

force in horizontal direction is given by F_x = 750 sin(23.9)

= 303.859 N

force in the vertical direction is F_y = 750 cos(23.9)

= 685.69 N.

Force is F = F_x i + F_y j

F = 303.859 i + 685.69 j N

(b)

Net force, F_net = F + F _{due to weight}

= 303.859 i + 685.69 j - 820j

= 303.859 i + 134.11 j N

(c)

Magnitude of F_net = ( F_x² + F_y² )^1/2

= ( 303.859² + 134.11² )^1/2

F_net = 332.138 N

(d)

angle = tan^-1 ( F_y / F_x )

= tan^-1 ( 134.11/ 303.859 )

= 23.81^o

(e)

F = ma

a = F / m

= 801.78 / g

a = 9.58 m/sec²

(f)

so the angle of acceleration is angle of force

= - 23.81^o