A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground
, the crane operators apply brakes to slow the unwinding motion. Knowing that the deceleration of cable A is 20 ft/s2 and the deceleration of cable B is 2 ft/s2, determine the tension in each cable.
1 answer:
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
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A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 *
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm