Tarzan, who weighs 649 N, swings froma cliff at the end of a convenient vine tha

Question

Tarzan, who weighs 649 N, swings froma cliff at the end of a convenient vine that is 20 m long (Figure 8-37). From the top of the cliff tothe bottom of the swing, he descends by 3.2 m. The vine will breakif the force on it exceeds 950 N. What is the greatest force on the vine during theswing?
Tarzan, who weighs 649 N, swings froma cliff at the end of a convenient vine that is 20 m long (Figure 8-37). From the top of the cliff tothe bottom of the swing, he descends by 3.2 m. The vine will breakif the force on it exceeds 950 N. What is the greatest force on the vine during theswing?

Explanation / Answer

I had the same question but with diff numbers.This is how u do it. Tarzan, who weighs 688 N, swings from a cliffat the end of a convenient vine that is 18 m long. From the top ofthe cliff to the bottom of the swing, he descends by 3.2 m. Thevine will break if the force on it exceeds 950 N. (a) Does the vinebreak? (b) If no, what is the greatest force on it during theswing? If yes, at what angle with the vertical does it break? When Tarzan jumps of the cliff his speed willincrease, until he reaches the bottom of the swing, after which itwill decrease. His speed leads to a centripetal force which adds tohis weight.
The speed and therefore the extra force will be maximum at thebottom of the swing, when both are pointing in the samedirection.

First we calculate the speed at the bottom of the swing. We put
m g h = (1/2) m v^2 and find v = sqrt ( 2 g h) = sqrt (2x9.813x3.2)= 7.92 m/s.
This speed adds a centripetal force of m v^2 /r to Tarzan'sweight.
F_centri = (688/9.813) 62.8 / 18 = 244 N.
Adding this to 688 N leads to 932 N. at the bottom of theswing.
So the vine doesn't brake. Tarzan's high speed brain did thecalculation before he jumped of course. When Tarzan jumps of the cliff his speed willincrease, until he reaches the bottom of the swing, after which itwill decrease. His speed leads to a centripetal force which adds tohis weight.
The speed and therefore the extra force will be maximum at thebottom of the swing, when both are pointing in the samedirection.

First we calculate the speed at the bottom of the swing. We put
m g h = (1/2) m v^2 and find v = sqrt ( 2 g h) = sqrt (2x9.813x3.2)= 7.92 m/s.
This speed adds a centripetal force of m v^2 /r to Tarzan'sweight.
F_centri = (688/9.813) 62.8 / 18 = 244 N.
Adding this to 688 N leads to 932 N. at the bottom of theswing.
So the vine doesn't brake. Tarzan's high speed brain did thecalculation before he jumped of course. When Tarzan jumps of the cliff his speed willincrease, until he reaches the bottom of the swing, after which itwill decrease. His speed leads to a centripetal force which adds tohis weight.
The speed and therefore the extra force will be maximum at thebottom of the swing, when both are pointing in the samedirection.

First we calculate the speed at the bottom of the swing. We put
m g h = (1/2) m v^2 and find v = sqrt ( 2 g h) = sqrt (2x9.813x3.2)= 7.92 m/s.
This speed adds a centripetal force of m v^2 /r to Tarzan'sweight.
F_centri = (688/9.813) 62.8 / 18 = 244 N.
Adding this to 688 N leads to 932 N. at the bottom of theswing.
So the vine doesn't brake. Tarzan's high speed brain did thecalculation before he jumped of course.

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