Vectors

[mykonos]

Dave and Jon kick a soccer ball at the same time. Dave kicks it with force of 120 N at angles of 60° and Jon kicks it with force of 200 N at angle of 120°. The angles are measured from a line between the centres of the two goals. Calculate the magnitude and direction of the resultant force.

Magnitude:
R = ?(200)^2 + (120)^2 - 2(200)(120)cos120
R = 280 N

I got magnitude, just need help solving direction please!

Spoiler: :3f2ruizr

280 N in a direction making an angle of 98.2°[/spoiler:3f2ruizr]

 

[DrLee]

Dave and Jon kick a soccer ball at the same time. Dave kicks it with force of 120 N at angles of 60° and Jon kicks it with force of 200 N at angle of 120°. The angles are measured from a line between the centres of the two goals. Calculate the magnitude and direction of the resultant force.

Magnitude:
R = ?(200)^2 + (120)^2 - 2(200)(120)cos120
R = 280 N

I got magnitude, just need help solving direction please!



Draw a picture of the two goals at the west and east end of the soccer field.

Dave's kick can be decomposed into two vectors as follows:
120 N sin(60) to the north and
120 N cos(60) to the east

Jon's kick can be decomposed into two vectors as follows:
200 N sin(120) to the north and
200 N cos(120) to the east

Now we can add the north vectors to get 320 N sqrt(3)/2 north = 277 N north
And we can see that the east vectors sum to -40 N east = 40 N west

So the resultant vector angle is the inverse tangent of 40/277 = 8.2 degrees north west. Since north is 90 degrees, the resultant is at an agle of 98.2 degrees.

You already have the magnitude of the vector, but you can also compute it as sqrt(40^2 + 277^2).

 

[mykonos]

Thank you that was very helpful.

"You already have the magnitude of the vector, but you can also compute it as sqrt(40^2 + 277^2)."

I didnt do that since my diagram did not show that there was a right angle that's why i used cosine law, instead of pythagorean theorem. am i wrong? you do get the same answer however.

 

[DrMike]

I didnt do that since my diagram did not show that there was a right angle that's why i used cosine law, instead of pythagorean theorem. am i wrong? you do get the same answer however.

According to the cosine law, if a triangle has edges length a,b,c, opposite angles A,B,C, then

\(\displaystyle a^2=b^2+c^2-2bc\cos A\)

If A is a right angle, then cos(A) = 0, so this becomes

\(\displaystyle a^2=b^2+c^2\)

So pythagoras' theorem is a special case of the cosine law. If you forget pythagoras, and stick to the cosine law forevermore, you'll never go wrong[sup:3negwfj7]1[/sup:3negwfj7], although you may end up doing a little extra work.
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[sup:3negwfj7]1[/sup:3negwfj7] That is, until you start working in spherical or hyperbolic space.