Cockroaches fighting phys. prob

[irefay]

I have an interesting problem for phys. tomorrow. I was hoping I might get a bit of help:

Massive cockroaches are fighting over a watermelon. Fred the Fierce is pulling on it with 310 Newtons at 45 degrees East of North. Larry the Large is pulling with 250 Newtons at 190 degrees from North. "Mandibles" Marlin is pulling at 285 degrees from North with a force of 275 Newtons.
a. Find the North/South and East/ West components of each vector. Use the appropriate cardinal points as directions (ex. a 5 Newton force applied southwest has components of 3.53 Newtons South and 3.53 Newtons west)

Force Choose 1 Force Choose 1
Fred:______Newtons (north or south)_______Newtons (East or west)

Larry:______Newtons (north or south)_______Newtons (East or west)

Marlin:_____Newtons (north or south)_______Newtons (East or west)

b. Find the resultant (magnitude and direction) when vectors Fred, Larry, and Marlin are added.


Well, Even if you can figure it out you should have a good laugh :)

 

[BobG]

That's just wrong. When the cockroaches get that strong it's time to forget about the exterminator and just torch the house.

Set one direction as positive along the x-axis (North, for example), which makes the opposite direction negative. Do the same for the y axis.

Once you've defined your positive negative directions, it's just a matter of breaking each force into its x component and y component. You do that using cosines and sines. Which you use for each component and vector depends on what directions you chose for positive x and positive y. If you draw out your vectors it should be pretty easy to picture which you need.

You add the vectors together using vector addition. In other words, all the x components are added together and all the y components are added together.

Using the pythagorean theorem and either the tangent, cosine, or sine, you can figure out part b.

 

[prasanna]

Solving the problem I get

a)Fred : 219.2 N (North) 219.2 N (East)
Larry: 246.2 N (South)43.412 N(West)
Marlin:71.175 N(North)265.63 N(West)

b)Adding up components in the y axis
F(y) = (219.2 + 71.175 - 246.2)N
= 44.175 N
Adding up the components on x axis
F(x) = (219.2 - 43.412 - 265.63) N
= -89.842 N
Magnitude = sqrt [ {F(x)}^2 + {F(y)}^2 ]
= 100.115 N
For finding direction,
tan A = F(y) / F(x)
= -0.4917
A = -26.18 degrees

Therefore the resultant is a force of magnitude 100.115 N
acting 26.18 degrees south of east.