a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc
1 answer:
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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
