Tarzan, who weighs 790 N, swings from acliff at the end of a 39.0 m vine thathan

Question

Tarzan, who weighs 790 N, swings from acliff at the end of a 39.0 m vine thathangs from a high tree limb and initially makes an angle of24.0° withthe vertical. Assume that an x axis extends horizontallyaway from the cliff edge and a y axis extends upward.Immediately after Tarzan steps off the cliff, the tension in thevine is 600 N. (a) At the moment of takeoff, what is the forceon Tarzan from the vine, in unit-vector form?
Tx = 1 N Ty = 2 N
(b) At that moment, what is the net force on Tarzan, inunit-vector form?
Fnet,x = 3 N Fnet,y = 4 N (c) What is the net force on Tarzan, as amagnitude and angle relative to the positive direction of thex-axis?
magnitude = 5 N direction = 6°
(d) What is the acceleration of Tarzan, as a magnitude anddirection, during takeoff?
magnitude = 7 m/s2 direction = 8°
(a) At the moment of takeoff, what is the forceon Tarzan from the vine, in unit-vector form?
Tx = 1 N Ty = 2 N
(b) At that moment, what is the net force on Tarzan, inunit-vector form?
Fnet,x = 3 N Fnet,y = 4 N (c) What is the net force on Tarzan, as amagnitude and angle relative to the positive direction of thex-axis?
magnitude = 5 N direction = 6°
(d) What is the acceleration of Tarzan, as a magnitude anddirection, during takeoff?
magnitude = 7 m/s2 direction = 8°
Tx = 1 N Ty = 2 N

Explanation / Answer

c) magnitude of the net force isFnet=[(244)^2+(-244.041)^2]=(59536+59556.009681)=345.10N now direction of force is given by tan=-(244.041N)/(244N) work it from here then the answer you'll get here will be the direction of the netforce below the horizontal d) mass of tarzan is w=m/g=790N/9.8m/s^2=80.61kg the magnitude of acceleration of tarzan isanet=Fnet/m=345.10N/80.61kg=4.281m/s^2 the direction of net acceleration is the same as your answer forthe direction of net force

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