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2 years ago

.Block A of mass 30 kg is resting on Block B of 15 kg and both blocks are connected via a

Engineering

1 answer:

Gala2k [10]2 years ago

6 0

The tension in the cable and the net acceleration of the block B are 291.814 newtons and 0.112 meters per square second, respectively.

<h3>How to determine the tension and the acceleration of block B</h3>

Based of the representation seen in the image, an external force moves the block A downwards and the block B moves upwards and with an net acceleration of same magnitude due to the cable. By Newton's laws we construct the free body diagrams for each mass and that can be seen in the image seen below.

The motion equations are described below:

Mass A

250 N - 0.4 · Na - T + (30 kg) · (9.807 m / s²) · sin 30° = (30 kg) · a (1)

Na - (30 kg) · (9.807 m / s²) · cos 30° = 0 (2)

Mass B

- T + 0.4 · Na + 0.3 · Nb + (15 kg) · (9.807 m / s²) · sin 30° = - (15 kg) · a (3)

Nb - Na - (15 kg) · (9.807 m / s²) · cos 30° = 0 (4)

By (2):

Na = (30 kg) · (9.807 m / s²) · cos 30°

Na = 254.793 N

By (4):

Nb = 254.793 N + (15 kg) · (9.807 m / s²) · cos 30°

Nb = 382.190 N

Then, we have the following system of linear equations:

T + (30 kg) · a = 250 N - 0.4 · Na + (30 kg) · (9.807 m / s²) · sin 30°

T + 30 · a = 295.188

T - (15 kg) · a = 0.4 · Na + 0.3 · Nb + (15 kg) · (9.807 m / s²) · sin 30°

T - 15 · a = 290.127

The solution of the linear system is T = 291.814 N and a = 0.112 m / s².

To learn more on Newton's laws: brainly.com/question/27573481

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Vsevolod [243]

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$\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}$

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Finally, to determine the normal stress in aluminum rod:

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3 years ago

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Answer:

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Explanation:

The parameters given are;

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Assumption,

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Steady state heat transfer process

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