The Tug will pull at all it's might to move the ice-slab,
Blue Ocean TowingOur company has contracted to protect oil platforms in the Davis Straight northeast of Canada from ice hazzards. A massive slab has cleaved and is moving with the sea current toward a production rig. Our biggest ocean tug will attach lines to the slab then applying a constant force perpendicular to the drift, to pull the estimated 800 x 106 kilogram aside so it passes no closer than four kilometers abreast of the rig. Use the geometry of the situation with the initial conditions. Solve Newton's Second Law to determine the least Constant Towing Force required of the ocean tug.
♦ The sketch (right) shows the slab, tug and rig at the commencement of the pull. To apply Newton's Second Law, we need a coordinate system with which to represent the vectors: force, position, and velocity. We place the origin at the rig with coordinates (and unit vectors I and J) parallel and normal to the flow. Two times of importance are the initiation of the tow (t = 0+) and the time when the slab is safely past the platform (t = tsafe). Since a force is to be calculated, the momentum equation with the slab, a mass of ice, and a few feet of the towing line are selected as system. The system boundary encompasses the slab, some line and cuts through the line. It is over the cut cross-section of the rope that the towing force acts. Elsewhere on the system boundary drag forces occur. These are proportional to velocity of the slab relative to the water. The slab floats with the water flow; drag is negligibly small.
To procede we integrate the momentum equation twice.
The first integration of the differential of momentum, is indefinite, and yields the upper limit, velocity, as a function of time.
Next, the velocity of the slab, V, is represented as the derivative of its vector position, dP/dt. Variables are separated and a second integration is performed. Notice the vector limits of the integral of position. The slab is initially directly upstream of the rig, and finally it is directly abreast of the rig.
The equation, scalar multiplied by J yields:
Next we consider the I component which states:
