How Is Tension Calculated in a Submerged Object's Lifting Cable?

[kiwikahuna]

Homework Statement

A seagoing crane lifts a sealed safe from a sunken ship. The average density of the safe is 5.3 grams/cc and its mass is 120 kg. What is the tension in the lifting cable when the safe is still submerged?

Homework Equations

Pressure = density *gravity* height*
Buoyant force = mg

The Attempt at a Solution

I calculated 5300kg/m^2 * 9.81 m/s^2 * 120 kg

The above is what I got so far but I'm not sure I'm doing it right. Any help would be greatly appreciated.

 

[CaptainZappo]

Draw a free-body diagram of the object. From the context of the problem, there must be no net force acting on the object. What forces act in which directions?

Note: the buoyant force is equal to the weight of the DISPLACED fluid, not necessarily equal to the weight of the submerged object.

 

[m2003]

I would suggest using the equation for buoyant force, Fb = ρVg, where ρ is the density of the fluid (in this case, water), V is the volume of the object (in this case, the safe), and g is the acceleration due to gravity. This equation takes into account the weight of the displaced fluid, which is equal to the buoyant force acting on the object.

Since the safe is submerged, the buoyant force acting on it is equal to its weight, which can be calculated using its density and volume. The volume can be found by dividing the mass by the density, giving a volume of approximately 22.6 liters (or 0.0226 m^3).

Thus, the buoyant force is equal to (5.3 g/cm^3) * (0.0226 m^3) * (9.81 m/s^2) = 1.18 x 10^3 N.

Since the safe is not accelerating, the tension in the lifting cable must be equal to the buoyant force. Therefore, the tension in the lifting cable would be approximately 1.18 x 10^3 N.