Anna notices a pyramid sticking out of an adjacent ravine and she wants to investigate. She notices some vines hanging over the ravine. To her luck, a strong breeze blows some of the vines toward her and she is able to grab one. The vine appears to make 10° angle with the vertical. She is curious to how long the ravine is so she can make a map of the region later on. (Anna's mass is 48kg.)

Question

___L /|\ L___

(a) Based on the location of the vine, which is right over the middle of the ravine, how wide is the ravine in terms of L?
(b) What she decides to do is tie a rock to the end of the vine and release it like it’s a pendulum. She decides she is going to time the swing to determine the period of the pendulum and from that she could figure out the length of the vine. She counts 10 cycles in 48.17 s. How wide is the ravine? (Hint: Look up the period for a pendulum)
(c) This is it; she has determined the width of the ravine and decides to swing across with out a running start. However, she forgot one thing, centripetal force. It might support her weight, but the vine might not be strong enough to support the swing. What Anna doesn’t know is the vine can only support a tension force of 480 N before it breaks. Will she make it to the other side or fall into the ravine? Prove your answer by finding the maximum tension in the vine.

work:
a) would it be cos(10)=x/L. I'm not sure how to answer this in terms of L.
b) not sure how to attempt. period of pendulum, 1.25?
c) Anna's mass is 48 kg so times that to 9.8 equals 470.4 so the tension must be greater than 470.4N?

Any form of help or guidance will be greatly appreciated. Thank you.

Answer 1

sin 10 = half width/L
so
half width = L sin 10 = .174 L
and'
width = .347 L
or
width/L = .347
----------------------------
only a chicken would look up pendulum period. We are tough.
x = A sin 2pi t/T
v = A(2 pi/T) cos 2 pi t/T
a = -A (2 pi/T)^2 sin 2 pi t/T
or a = -(2 pi/T)^2 x
so
a = - C
F = -m g x/L = m a
so
-m g x/L = - m (2 pi/T)^2 x
or
g/L = (2 pi/T)^2
here we measured T = 4.817 seconds
so
9.81/L = (2 pi/4.18)^2
L = 4.34 meters
so
width = .347 L = 1.51 meters

Answer 2

You are right but we need to find that centripetal acceleration to really answer part c
find max velocity
v = A(2 pi/T) cos 2 pi t/T
max of cos is 1 in the middle of the ravine of course
A = half width of ravine
= 1.51/2 = .755 meter
so vmax = .755(2 pi/4.817)
= .985 meters/second
centripetal =Ac = v^2/r
= .985^2/4.34
= .223 m/s^2
so max tension = ma + mg
= 48( .223 + 9.81)
=482 Newtons
that is a little more than we need to break the vine

Answer 3

Thank you so much Damon! I really appreciate it!! I could not understand any of this without your help!