Statics: 2-D Resultants

We're now looking at summing the forces to get the resultant force which is the single force that we replace all the others with and the system will still be under the same force and respond the same way we would have under all individual forces. We can do the same thing for the moments. 

Those forces contribute to both the force and the moment. You change both of the sum if you change both the forces and the moments. 

There's something associated with the sum of the moment, it's like a steering wheel. Think about grabbing the wheel and turn it, and that's this business about the sum of the moment. We can get a wheel at 2 different places and giving it a turn in a counterclockwise direction, and in general those are about the same size. The reason the wheels are so big is to give a nice big moment arm so that it's very easy to exert a lot of torque even with a small amount of force. 

The moment is taken about the center point and it is just 2Fd because there are 2 equal and opposite forces. 2 equal and opposite forces separated by some distance (2D) is known as a couple. Couples are 2 moments by equal and opposite given the symbol C as the magnitude of the forces times the distance between them.  C = 2Fd. This is equal and opposite forces separated by some distance. We need to figure out the force vectors exerted by those tugs and sum those 2 forces up and calculate the moment each one of them exerts, causing the ship to turn a little bit. We'll get to the last part about the super tug as well. 

We're working in a 2d problem since all the moment is going to be in the k direction since it turns clockwise or counterclockwise. We do R x F for each one of the forces, or break each force into the vector components, which is the cross product. 

We have one force as an example now: 

Let's say F2 = 3000i - 4000j. 

Say F3 = -5000j lb

and F4 = (3540i + 3540j ) lb.

Now, time to figure out the moment of each of the forces. 

M1 = (r1 x F1) about an axis. How would we sketch r1? 

We draw from the point o to the line of action where the boat touches. 

Let's say x distance is 90 ft and Y distance is 40 ft. Then here's the matrix: 

There should be no X and Y components for cross product. 

As a result, we get i(0) - j(0) - k(-90(-4330) - (50)(2500))

which is equivalent to (265000 lb ft) k. 

The k direction is actually the cross  product of i and j. 

We get the second moment to be (-610,000 lb ft) k. M3 = -2,000,000 lb ft k and M4 = 1,310,000 lb ft k.

So  here's the final equations: 

And you can get the moment, get the force, and look for the supertug.