# Two tugs are towing a ship. The smaller tug is 15 degree off the port bow, and the larger tug is 20 degree off the starboard bow. The larger tug pulls twice as hard as the smaller tug. In what...

Two tugs are towing a ship. The smaller tug is 15 degree off the port bow, and the larger tug is 20 degree off the starboard bow. The larger tug pulls twice as hard as the smaller tug. In what direction will the ship move?

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To understand the answer to this question on must understand how to add force vectors in a coordinate plane. Although this can be done graphically by making a scale drawing and representing the size and direction of the vectors as arrows, it is probably easier and quicker to add the vectors together using basic trigonometry.

To do this one needs to remember how to break a vector into its horizontal and vertical components. Once that is done for each vector, the next step is to add the vertical components for each vector together and then to add the horizontal components together. The resulting two answers will be the two legs of a right triangle. The direction and size of the total force will be determined by the hypotenuse of that triangle.

To begin we will first define "port" as negative and "starboard" as positive thus creating the two vectors

F1 = F at -15 deg

F2 = 2F at 20 deg

The horizontal components are

F1x + F2x = Fcos(-15)+2Fcos(20) = F[cos(-15) + 2cos(20)] = 2.85F

The vertical components are

F1y + F2y = Fsin(-15) + Fsin(20) = F[sin(-15) + 2sin(20)] = 0.425F

By Pythagorus' Theorem

`"net" F =sqrt((2.85F)^2 +(0.425F^2))`

`"net"F = sqrt(8.30F^2) = 2.88F`

To get the angle we use

`theta ="tan"^(-1)((0.425F)/(2.85F)) = 8.48 deg`

The result of the two tugs pulling against the ship will be that the ship will be pulled with a force 2.88F

**in a direction of 8.48 degrees to the starboard bow.**