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koban [17]
3 years ago
8

Block A has a weight of 8 lb. and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic fric

tion is 0.2 . If the spring between them has a stiffness of k=20 lb/ft, and it is compressed 0.2 ft, determine the acceleration of each block just after they are released.

Engineering
1 answer:
kkurt [141]3 years ago
4 0

Answer:

For block A, a = 9.66 ft/s²

For block B, a = 15 ft/s²

Explanation:

A free body diagram for this force system is attached to this solution

Mass of block A = m₁ = 8 lb

Mass of block B = m₂ = 6 lb

Coefficient of kinetic friction = μ

Normal reaction on the blocks = N

Spring stiffness of the spring btw block A and B = k = 20 lb/ft

Compression of the spring = 0.2 ft

Analysing Block A first

The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 8 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 8/32.2 = 0.248 lbm

(20×0.2) - (0.2 × 8) = (0.248) aₓ

aₓ = 9.66 ft/s²

The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 6 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 6/32.2 = 0.186 lbm

(20×0.2) - (0.2 × 6) = (0.186) aₓ

aₓ = 15 ft/s²

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Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603                         225

0.203                         368

0.162                         442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

Bending Stress = \frac{1}{(Mirror Radius)^{n}}

At mirror radius 1 = 0.603 mm   Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm  Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm   Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}

\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}

0.6114 = (0.3366)^{n_1}

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}

\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}

0.8326 = (0.7980)^{n_2}

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

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comparing case 1 and 3 using the above equation

\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}

\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}

0.5090 = (0.2687)^{n_3}

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

n = \frac{n_1 + n_2 + n_3}{3}

n = \frac{0.452 +0.821 + 0.514}{3}

n = 0.596

Hence to get bending stress x at mirror radius 0.796

\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}

\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}

\frac{Stress x}{225} = 0.8475

stress x = 191 MPa

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Answer:

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Explanation:

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Take C_A as point  on which nitrogen concentration is 2 kg/m^3

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Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

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