Walking by a pond, you find a rope attached to a tree limb 5.2 m off the ground.
ID: 1696629 • Letter: W
Question
Walking by a pond, you find a rope attached to a tree limb 5.2 m off the ground. You decide to use the rope to swing out over the pond. The rope is a bit frayed but supports your weight. You estimate that the rope might break if the tension is 84 N greater than your weight. You grab the rope at a point 4.6 m from the limb and move back to swing out over the pond. Assume your weight to be 650 N (about 145 lb).(a) What is the maximum safe initial angle between the rope and the vertical so that it will not break during the swing?
1°
(b) If you begin at this maximum angle, and the surface of the pond is 1.2 m below the level of the ground, with what speed will you enter the water if you let go of the rope when the rope is vertical?
2 m/s
Explanation / Answer
(a)
Use conservation of energy
Wext = K + U = 0
or
K2 - K1 + U2 - U1 = 0
But K1 = 0
So (1/2)mv2^2 + mgh - [mgL(1 - cos) + mgh] = 0
On multiplying the the above equation with (1/mg) we get
v2^2/2g - L + L cos = 0
L cos = L - v2^2/2g
cos = 1 - v2^2/2gL
= Cos^-1[1 - (v2^2/2gL)]
From newton's second law
T - mg = mv2^2/L
and T = mg + mv2^2/L
Given that rope might be break if the tension in it exceeds weight by 80 N so
mv2^2/L = 84 N
v2^2 = 84L/m
If your weight is 650 N,then mass m = 66.3 Kg
v2^2 = 84*4.6/66.3 = 5.828 m^2/s^2
Then = Cos^-1[1 -(5.828/2*9.8*4.6)]
= 20.71 degrees
(b)
Wext = K + U = 0
K3 - K1 + U3 - U1 = 0 where U3 and K1 are Zero.
(1/2)mv3^2 - mg[h + L(1-cos) = 0
v3 = Sqrt[2g[h + L(1- cos)]
v3 = Sqrt[2*9.8[1.8 + (4.6)(1-cos20.71)]
= 6.41 m/s
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