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Free_Kalibri [48]
3 years ago
10

The two boxcars A and B have a weight of 20 000 Ib and 30 000 Ib, respectively. If they coast freely down the incline when the b

rakes are applied to all the wheels of car A causing it to skid, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is μk=0.5. The wheels of car B are free to roll. Neglect their mass in calculation.

Engineering
2 answers:
harkovskaia [24]3 years ago
8 0

Answer:

a = 3.608 ft/s²

T = 5.977 kip

Explanation:

Angle of inclined track is not given. It is 5 degree as shown in attached image 1.

Given data:

Weight of boxcar A = Wα = 20,000 lb

Weight of boxcar B = Wb = 30,000 lb

Coefficient of kinetic friction = υk = 0.5

angle of inclined track = 5 degree

Value of g = 32.2 ft/s²

Solution:

These boxcars are moving therefore,

∑Fₓ = ma

∑Fy = 0

Car A

Use ∑Fy = 0

Forces acting in Y-axis are weight and its counter N (as shown in Free body diagram). Weight is Negative (acting downward) while N is positive (acting upward)

∑Fy = Nα - Wα*Cos(∅) = 0

∑Fy = Nα - 20,000*Cos(5°) = 0

∑Fy = Nα - 20,000*0.9962 = 0

∑Fy = Nα - 19,924 = 0                

Nα = 19,924

Use ∑Fₓ = maₓ

Forces acting in x-axis are Fa, T and horizontal component of weight and force having acceleration a (as shown in Free body diagram). Horizontal component of weight and Tension (T) are Negative (acting in negative X-axis direction) while Fa is positive (acting in positive X-axis direction)

∑Fₓ = maₓ

(0.5*19924) - 20,000*Sin(5°) - T = (20,000/32.2)*a

9962 - 1744 - T = 621.12*a

8218 - T = 621.12*a  .................. Eq(1)

Both cars

∑Fₓ = maₓ

Forces acting are; horizontal component of weights, Fa and force having acceleration a (as shown in Free body diagram).

(0.5*19924) - 20,000*Sin(5°) - 30,000*Sin(5°) = (20,000/32.2)/a + (30,000/32.2)/a

9962 - 1744 - 2616 = 621.12/a + 931.68/a

5602 = 1552.8 / a

⇒ a = 3.608 ft/s²

put a = 3.608 ft/s² in equation (1)

⇒ T = 8218 - 621.12*3.608

    T = 8218 - 2240.79

    T = 5977.21 lb

    T = 5.977 kip

Tpy6a [65]3 years ago
3 0

Answer:

Answer for the question :

"the two boxcars A and B have a weight of 20 000 Ib and 30 000 Ib, respectively. If they coast freely down the incline when the brakes are applied to all the wheels of car A causing it to skid, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is μk=0.5. The wheels of car B are free to roll. Neglect their mass in calculation."

is explained in the attachment.

Explanation:

Download pdf
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Convert 25 mm into in.
mihalych1998 [28]

Answer:

25 mm = 0.984252 inches

Explanation:

Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:

<u>1 mm = 1/25.4 inches</u>

From the question, we have to convert 25 mm into inches

Thus,

<u>25 mm = (1/25.4)*25 inches</u>

So,

25 mm=\frac{25}{25.4} inches

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3 years ago
The correct statement about the lift and drag on an object is:_______
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Answer:

(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift

Explanation:

When a fluid flows around the surface of an object, it exerts a force on it. This force has two components, namely lift and drag.

The component of this force that is perpendicular (normal) to the freestream velocity is known as lift, while the component of this force that is parallel or in the direction of the fluid freestream flow is known as drag.

Lift is as a result of pressure differences, while drag results from forces due to pressure distributions over the object surface, and forces due to skin friction or viscous force.

Thus, drag results from the combination of pressure and viscous forces while lift results only from the<em> pressure differences</em> (not pressure forces as was used in option D).

The only correct option left is "A"

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Read 2 more answers
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

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5 0
3 years ago
Water is being added to a storage tank at the rate of 500 gal/min. Water also flows out of the bottom through a 2.0-in-inside di
melomori [17]

Answer:

From the answer, the water level is falling (since rate of outflow is more than that of inflow), and the rate at which the water level in the storage tank is falling is

(dh/dt) = - 0.000753

Units of m/s

Explanation:

Let the volume of the system at any time be V.

V = Ah

where A = Cross sectional Area of the storage tank, h = height of water level in the tank

Let the rate of flow of water into the tank be Fᵢ.

Take note that Fᵢ is given in the question as 500 gal/min = 0.0315 m³/s

Let the rate of flow of water out of the storage tank be simply F.

F is given in the form of (cross sectional area of outflow × velocity)

Cross sectional Area of outflow = πr²

r = 2 inches/2 = 1 inch = 0.0254 m

Cross sectional Area of outflow = πr² = π(0.0254)² = 0.00203 m²

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Rate of flow of water from the storage tank = 0.0203 × 18.288 = 0.0371 m³/s

We take an overall volumetric balance for the system

The rate of change of the system's volume = (Rate of flow of water into the storage tank) - (Rate of flow of water out of the storage tank)

(dV/dt) = Fᵢ - F

V = Ah (since A is constant)

dV/dt = (d/dt) (Ah) = A (dh/dt)

dV/dt = A (dh/dt) = Fᵢ - F

Divide through by A

dh/dt = (Fᵢ - F)/A

Fi = 0.0315 m³/s

F = 0.0371 m³/s

A = Cross sectional Area of the storage tank = πD²/4

D = 10 ft = 3.048 m

A = π(3.048)²/4 = 7.30 m²

(dh/dt) = (0.0315 - 0.0370)/7.3 = - 0.000753

(dh/dt) = - 0.000753

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