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Charra [1.4K]
3 years ago
7

A person is planning a bungee jump from a 40 meter high bridge. Under the bridge is a river with crocodiles, so the person does

not want to be submerged into the water. The rubber rope fastened to the ankles has a spring constant of 3,600 N/m per meter length. Thus one meter length of rope will have a spring constant of 3,600 N/m, two meters of rope will have a spring constant of 1,800 N/m, three meters of rope will have a spring constant of 1,200 N/m and so forth. The distance from the ankles to the top of the head is 175 cm, and the person has a mass of 80 kg. Calculate: a. b. c. how long the rope should be [m] how much deflection the rope experiences [m] how much impact force the bungee jumper experiences [N]
Engineering
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

a. l = 19.7m, b. 18.55m, c. Impact Force = 3889.84 N

Explanation:

The total energy of the system when the person is at top of the bridge is

Potential energy = mgh, Kinetic energy = 0

The total energy of the  system when the person reaches just above the surface

Potential energy = 0, Kinetic energy = 0, Spring energy = ½ K X2, where k is the spring constant and X is the deflection

Applying conservation of energy

mgh = 0 + 0 + ½ K X²

80 x 9.81 x 40 = ½ (3600/l) X²

31392 = ½ (3600/l) X²

We can also conclude that

l+ X + 1.75 = 40

l + X = 38.25

a. <u>Substitute the value of x from above into the energy conversion expression</u>

31392 = ½ (3600/l)(38.25 - l)²

31392 x 2/3600 = (38.25 + l² – 2l(38.25))/l

17.44l = l2 – 76.5l + 38.25²

l² – 76.5l – 17.44l +1463.0625 = 0

Solving for l we get

L = 19.7

Hence, length of the rope is 19.7m

b. <u>The deflection is calculated by using the relation between l and X</u>

L + X = 38.25

X = 38.25 – 19.7 = 18.55m

c. <u>The impact force is calculated using the impact force formula which relates the impact force with the deflection</u>

F = KX

F = (3600/l) . X

F = (3600/19.7) . (18.55) = 3889.84 N

Thus, the impact force is 3889.84 N

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3 years ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
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Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

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Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

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k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
3 years ago
How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.
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Answer:

N_3=\dfrac{T_1}{T_3}N_1

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

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Number\ of\ teeth\ of\ gear 1=T_1

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So the speed of gear third can be given as follows

\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}

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