Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 134P

(a)

To determine

The tension in the vine just before it broke.

(a)

Expert Solution
Check Mark

Answer to Problem 134P

The tension in the vine just before it broke is 1.64kN.

Explanation of Solution

Find the equation for the length of the vine.

sinθi=oppositehypotenuse=5.00mLL=5.00msinθi (I)

Here, θi is the initial angle between vine and the vertical and L is the length of the vine.

Find the equation for the change in height.

Δy=yfyi=LcosθfLcosθi=L(cosθfcosθi) (II)

Here, θf is the final angle between vine and the vertical.

Use the law of conservation of energy to find the equation for the person T’s speed when the vine breaks.

ΔK+ΔU=012m(vf2vi2)=mgΔyvf2vi2=2g|Δy|vf0=2gL(cosθfcosθi)vf=2gL(cosθfcosθi) (III)

Here, mg is the weight of person T, g is the acceleration due to gravity, ΔK is the change in kinetic energy, ΔU is the change in potential energy, vi is the initial velocity of person T and vf is the final velocity of person T.

Find the equation for the tension.

ΣF=Tmgcosθfmvf2L=Tmgcosθfmgvf2gL=TmgcosθfT=mg(vf2gL+cosθf) (IV)

Conclusion:

Substitute 60° for θi in equation (I) to find L.

L=5.00msin60°=5.77m

Substitute 5.77m for L, 9.80m/s2 for g, 60° for θi and 20° for θf in equation (III) to find v.

vf=2(9.80m/s2)(5.77m)(cos20°cos60°)=7.05m/s

Substitute 5.77m for L, 9.80m/s2 for g, 7.05m/s for vf, 900.0N for mg and 20° for θf in equation (IV) to find T.

T=(900.0N)((7.05m/s)2(9.80m/s2)(5.77m)+cos20°)=1.64×103N=1.64kN

Thus, the tension in the vine just before it broke is 1.64kN.

(b)

To determine

If person T land on the other side of the river safely.

(b)

Expert Solution
Check Mark

Answer to Problem 134P

Yes, person T lands on the other side of the river safely.

Explanation of Solution

Find the distance to the ground level at the moment of the vine breaking.

Δy=8.00mLcosθf (V)

Find the equation for the distance to the river edge.

r=Lsinθf (VI)

Here, r is the distance to the river edge.

Find the quadratic equation in terms of time taken by person T to reach ground.

Δy=vsinθfΔt12g(Δt)212g(Δt)2+vsinθfΔt+Δy=0 (VII)

Find the equation for the horizontal distance travelled.

Δx=vcosθfΔt (VIII)

Here, Δx is the horizontal distance travelled.

Conclusion:

Substitute 5.77m for L and 20° for θf in equation (V) to find Δy.

Δy=8.00m(5.77m)cos20°=2.58m

Substitute 5.77m for L and 20° for θf in equation (VI) to find r.

r=(5.77m)sin20°=1.97m

Substitute 2.58m for Δy, 9.80m/s2 for g, 7.05m/s for vf and 20° for θf in equation (IV) to find T.

12(9.80m/s2)(Δt)2+(7.05m/s)sin20°Δt2.58m=04.9m/s2(Δt)2+(2.41m/s)Δt2.58m=0

Solve the quadratic equation to find the positive time interval.

Δt=b±b24ac2a=2.41m/s±(2.58m)2(4.9m/s2)=0.52s

Substitute  7.05m/s for vf, 20° for θf and 0.52s for Δt in equation (VIII) to find Δx.

Δx=(7.05m/s)cos20°(0.52s)=3.4m

As the distance travelled in this time is greater than the width of the river.

3.4m>1.97m

Thus, the person T lands on the other side of the river safely.

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Chapter 6 Solutions

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