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goldenfox [79]
3 years ago
15

What major advancement in machine tools occurred in the 1970s and what benefits did it provide? describe in your own words.

Engineering
2 answers:
mixer [17]3 years ago
4 0

Answer:

I'm just a seventh grader

kolbaska11 [484]3 years ago
4 0

Answer:

here's an essay I wrote one this topic

Explanation:

The early 1970s saw price and wage freezes and high unemployment. The oil crisis of 1973 and 1979 served to help slow an already snail-paced economy, and by 1979, the U.S. was experiencing the worst inflation in 33 years.

A growing world market was not helping the economic transition as traditionally strong big businesses like automotive and steel lost market shares to Europe and Japan. At the beginning of the decade, the government bailed out Lockheed Martin with $250 million worth of loan guarantees. Later it did the same for Chrysler—to the tune of $1 billion.

Obviously, the old ways of doing business were failing. The seeds of concern about more efficient operations, more flexibility, and higher quality were taking root in this decade for the eventual growth of a new paradigm in manufacturing that would flourish in the 80s. As NCs and DNCs gained recognition and popularity, the benefits of faster, more accurate cutting began to make demands on traditional tooling, controls, inspection, and other facets of the manufacturing process. This led to the development of many new companies who were ready to meet the challenges of a transitioning decade and make the products that would fulfill the needs of a new world of manufacturing.

As manufacturers looked for inexpensive solutions to their problems, those in the machine tool industry responded with new, lower-cost technologies. GE Fanuc Automation for one developed its Mark Century 550 Series in 1970. The highly-reliable control system was marketed as a complex, low-cost NC and became popular in the worldwide market. It introduced new features like lathe canned cycles, spindle speed monitor, battery backed tool offset, RAM memory, and cutter compensation.

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The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

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Answer:

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