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3 years ago

7

If the motor exerts a constant force of 30 lb on the cable, determine the power that must be supplied to the motor if the load h

as been hoisted s = 10 ft starting from rest. the motor has an efficiency of ϵ = 0.76.

1 answer:

valentinak56 [21]3 years ago

6 0

Try chegg they have answers

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Explanation:

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Answer:

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Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

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$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

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So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

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3 years ago

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3 years ago

A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d

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Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

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The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

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3 years ago