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3 years ago

5

A 14-lb crate is pulled up a frictionless 40° ramp with an initial velocity of v1=0.4 ft/s. It is pulled 0.3 ft from location #1

to #2 by a pulley system and a P=36-lb force pulling down on pulley A. Also, a spring (k=69 lb/ft) is attached
to the crate which is un-stretched at location #2. Assume the spring, cable, and pulleys are massless. Find the final velocity of the crate in ft/sat location #2.

1 answer:

Morgarella [4.7K]3 years ago

6 0

Answer:

3.25 ft/s

Explanation:

The crate is of =14-lb=m₁

The angle of inclination is = 40°=Ф

The initial velocity = 0.4 ft/s= v₁

Distance the crate will move is= 0.3 ft =d

The load pulling downwards is = 36 lb= m₂

Acceleration of the pulley, a= m₂g - m₁gsinФ / m₁+m₂ where g= 32.17 ft/s^2

a= 36*32.17 - 14*32.17*sin 40° / 14+36

a=17.37 ft/s^2

Apply the formula for final velocity

V₂²=V₁²+2ad

V₂²=0.4²+ 2*17.37*0.3

V₂²=10.582

V₂ =√10.582 = 3.25 ft/s

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3 years ago

You have just finished your OST takeoffs for a single-story home and found 175 LF of interior walls with 2x6 studs 14" OC. The h

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Answer:

Total BF for the interior wall is 7.50BD

Explanation:

Given Data:

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· Top plates = 2

· Bottom Plate = 1

BF stands for board feet in lumber/wood terminology. It is the unit of volume.

1 BF (Board feet) = 1 ft x 1 ft x 1 inch

Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.

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3 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

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Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

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Re = $\frac{\rho \times U\times x}{\mu }$

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= 1992316

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= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

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Answer:

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Explanation:

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minimum electric power consumption of the fan motor

solution

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E in - E out = $\frac{dE \ system}{dt}$ ......................1

and at steady state $\frac{dE \ system}{dt}$ = 0

so we can say from equation 1

E in = E out

so

minimum power required is

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put here value

E in = $\rho A V \frac{V^2}{2}$

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