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...

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.