Question

how much force was placed on my back. object lifted on a 25 lb anchor was a 300 yardlong power cable. the weight per foot of cable was 7.97 lbs. per linear foot. the cable was lifted 62 feet to the surface of the water. I am 5 feet 7 inches tall and it took me 4 hours to lift it unaided. how much force and how much weight was placed on my back.

Answer 1

How much force was placed on my back. object lifted on a 25 lb anchor was a 300 yardlong power cable. the weight per foot of cable was 7.97 lbs. per linear foot. the cable was lifted 62 feet to the surface of the water. I am 5 feet 7 inches tall and it took me 4 hours to lift it unaided. how much force and how much weight was placed on my back.

Solution:

$L = 300$ yards $= 274.32m -$ length of the cable

$M = 25lb = 11.3kg - mass$ of the anchor

$m = 7.97\frac{\mathrm{lb}}{ft} = 3.61\frac{\mathrm{kg}}{\mathrm{ft}} = 11.84\frac{\mathrm{kg}}{\mathrm{m}} -$ weight per foot of cable

$H = 62ft = 18.9m -$ lifting height of the cable

$h = 5$ feet 7 unches $= 1.7m -$ my height

Total weight to be lifted:

$$M _ {1} = M _ {\text {anchor}} + M _ {\text {cable}} = M + m \cdot L = 11.3 \mathrm{kg} + 11.84 \frac {\mathrm{kg}}{\mathrm{m}} \cdot 274.32 \mathrm{m} = 3259 \mathrm{kg}$$

Newton's second law along the Y-axis:

$$\vec {F} + \overrightarrow {M _ {1} g} = 0 \quad (a = 0, \text {because } V = \text {constant})$$ $$y: F = M_{1}g$$ $$F = 3259kg \cdot 9.8 \frac{N}{kg} = 31938.2N$$

Answer: $M = 3259kg; F = 31938.2N$