# Vector Problems

• Apr 3rd 2008, 06:27 AM
seagirt24
Vector Problems
Hi, I am having trouble with problems involving vectors:

A power boat crossing a wide river has a compass heading of 25 degrees and a speed relative to the water of 15 mph. The river is flowing in the direction of 135 degrees at 3.9 mph. What is the boat's actual velocity, that is, what is its speed and direction relative to the ground?

A large ship has gone aground in a harbor and two tugs, with cables attached, attempt to pull it free. If one tug pulls with a compass course of 161 degrees and a force of 2900 kg, and a second tug pulls with a compass course of 192 degrees and a force of 3600 kg., what is the compass direction and the magnitude of the resultant force?

I think I am setting both of these problems up right, and doing the math right, but I'm unsure of how to find the angles in the triangle created by the resultant vector. (Shake)

If some math genius could help me figure this out that would be amazing!!!!! (Cool)
• Apr 3rd 2008, 07:00 AM
earboth
Quote:

Originally Posted by seagirt24
Hi, I am having trouble with problems involving vectors:

A power boat crossing a wide river has a compass heading of 25 degrees and a speed relative to the water of 15 mph. The river is flowing in the direction of 135 degrees at 3.9 mph. What is the boat's actual velocity, that is, what is its speed and direction relative to the ground?

...

I've attached a more or less exact drawing of the situation.

Divide the parallelogram into 2 congruent triangles with 2 sides and the included angle given. Use Cosine rule to calculate the missing values.

cc = compass course
cog = course over ground
• Apr 3rd 2008, 08:18 AM
seagirt24
thanks... I think I'm clear on the picture. The thing I'm still confused about is, how do the angles given in the problem relate to the angles that are in the resultant triangle? e.g. how do the 25 and 135 factor into the triangle?
• Apr 3rd 2008, 02:15 PM
seagirt24
I know that there has to be a 65 degree angle in the triangle, but I don't know how to get it... any suggestions?
• Apr 3rd 2008, 11:39 PM
earboth
Quote:

Originally Posted by seagirt24
thanks... I think I'm clear on the picture. The thing I'm still confused about is, how do the angles given in the problem relate to the angles that are in the resultant triangle? e.g. how do the 25 and 135 factor into the triangle?

Since the boat is heading at 25° and the river flows at 135° the angle at the vertex of the parallelogram must be

$135^\circ - 25^\circ = 110^\circ$

Then the angle at the other vertex of the parallelogram must be

$180^\circ - 110^\circ = 70^\circ$

Now you can calculate the resulting diagonal (=cog) using the Cosine rule:

$|cog| = \sqrt{15^2+3.9^2-2\cdot 15 \cdot 3.9 \cdot \cos(70^\circ)} \approx 14.14898...$