- #1
MFlood7356
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1. Tarzan, who weighs 790 N, swings from a cliff at the end of a 24.2 m vine that hangs from a high tree limb and initially makes an angle of 26.3° 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 708 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?
2. F=ma Fx=ma-Fcos(theta) Fy=-Fsin(theta)
3. I really don't understand how to do any of these because the two forces and the length of the rope threw me off. I wasn't sure which force to use where. This is as much as I've done but I don't know what parts they relate to. I would really appreciate it if someone would help me:
Fx= 790N -708Ncos26.3= 155N
Fy = -708sin26.3= -314N
magnitude= SQRT1552+-3142= 350N
inversetan(-314N/155N)= -63.7 degrees
2. F=ma Fx=ma-Fcos(theta) Fy=-Fsin(theta)
3. I really don't understand how to do any of these because the two forces and the length of the rope threw me off. I wasn't sure which force to use where. This is as much as I've done but I don't know what parts they relate to. I would really appreciate it if someone would help me:
Fx= 790N -708Ncos26.3= 155N
Fy = -708sin26.3= -314N
magnitude= SQRT1552+-3142= 350N
inversetan(-314N/155N)= -63.7 degrees