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Tju [1.3M]
3 years ago
11

A cubic shaped box has a side length of 1.0 ft and a mass of 10 lbm is sliding on a frictionless horizontal surface towards a 30

upward incline. The horizontal velocity of the box is 20 ft/s. Determine how far up the incline the box will travel (report center of mass distance along the inclined surface, not vertical distance)
Engineering
1 answer:
natulia [17]3 years ago
6 0

Explanation  & answer:

Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.

Initially, the cube has a kinetic energy of

KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2  / 2 = 2000 lbm-ft^2 / s^2

At the highest point when the block stops, the gain in potential energy is

PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

By assumption, there was no loss in energies, we equate PE = KE

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

=>

h = 2000 /322 = 6.211 (ft)

distance up incline = h / sin(30) = 12.4 ft

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You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?
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Answer:

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Explanation:

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1. Which of the following is NOT an example of a lever?
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A staircase what makes it a lever is another objects used to displace the force better
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4 years ago
A journal bearing has shaft with a diameter of 75 mm. The bushing bore is 75.1 mm and the bushing is 37.5 mm long and supports a
Rainbow [258]

Explanation:

The given data is as follows.

   Length (l) = 37.5 mm,       Journal diameter (d) = 75 mm

   Diameter of bushing (D) = 75.1 mm,   Speed (N) = 720 rpm,

   Load (F) = 2 kN

Dynamic viscosity of SAE 20 at 60^{o}C is 12.2 cp

  Clearance = (\frac{D}{2}) - (\frac{d}{2})

           c = (\frac{75.1}{2}) - (\frac{75}{2})

              = 0.05 mm

Bearing pressure will be calculated as follows.

   Bearing pressure (p) = \frac{F}{l \times d}

                            = \frac{2 \times 10^{3}}{37.5 \times 75}

                            = 0.711 N/mm^{2}

Now, sommerfield number (S) will be calculated as follows.

      S = (\frac{r}{c})^{2} \times \frac{\mu \times N}{p}

          = (\frac{37.5}{0.05})^{2} \times \frac{12.2}{10^{9}} \times \frac{720}{60} \times \frac{1}{0.711}

          = 0.006

As,   \frac{l}{d} = \frac{37.5}{75} = \frac{1}{2}

Now, using Raimondi and John Boyd data the values for other variables will be obtained.

Minimum film thickness variable = (\frac{h_{o}}{c}) = 0.03

Therefore, minimum film thickness (h_{o}) = 0.03 \times 0.05

                                           = 0.0015 mm

As, (\frac{P}{P_{max}}) = 0.126

Hence, maximum lubricant pressure will be calculated as follows.

       P_{max} = \frac{0.711}{0.126}

                      = 5.642 N/mm^{2}

Due to friction, the heat loss rate will be as follows.

            \frac{2 \pi NfFr}{10^{6}} kW

According to coefficient of friction variable = \frac{r}{c} \times f = 0.61

              f = \frac{c}{r} \times 0.61

                = \frac{0.05}{37.5} \times 0.61

                = 0.000813

Therefore, heat loss will be calculated as follows.

      Heat loss = \frac{2 \pi \times \frac{720}{60} \times 0.000813 \times 2 \times 10^{3} \times 37.5}{10^{6}}

                      = 4.6 Watt

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