Answer: hello your question is incomplete below is the complete question
answer:
N010 GO2 X7.0 Y2.0 15.0 J2.0 ( option 1 )
Explanation:
Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path
GO2 = circular interpolation in a clockwise path
G91 = incremental dimension
<em>hence the correct option is </em>:
N010 GO2 X7.0 Y2.0 15.0 J2.0
Answer:
I. Tension (cable A) ≈ 6939 lbf
II. Tension (cable B) ≈ 17199 lbf
Explanation:
Let's begin by listing out the data that we were given:
mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,
g = 32.17405 ft/s²
The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.
Mathematically represented thus:
T = mg + ma
where:
T = tension, m = mass, g = gravitational force,
a = acceleration
I. For Cable A, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-20)]
T = 18339.2085 - 11400 = 6939.2085
T ≈ 6939 lbf
II. For Cable B, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-2)]
T = 18339.2085 - 1140 = 17199.2085
T ≈ 17199 lbf
Answer:
Hello your question is incomplete attached below is the complete question
Answer : Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Explanation:
Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Attached below is the detailed solution
