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GaryK [48]
3 years ago
12

A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value

of Y for which the system is critically damped. Assume that g=32 ft/s2. Round your answer to three decimal places.
Engineering
1 answer:
Dominik [7]3 years ago
5 0

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

m = \dfrac{22}{32}=0.6875\ s^2/ft

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

K=\dfrac{m}{x}=\dfrac{22}{0.375}

K= 58.66  lb/ft

The damper coefficient for critically damped system

C_c=2\sqrt{mK}

C_c=2\sqrt{0.6875\times 58.66}

Cc= 12.7 lb.sec/ft

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A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat
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no of unit is 17941

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