Forces on Particles: 3 of 4 Forces Acting on 10kg Mass

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Homework Statement



3 of 4 Forces acting on a 10kg mass in free space are shown in figure below.

a) Express these forces in unit vector notation

b) Find the magnitude and direction of the fourth force F4 so that the particle maintains a constant velocity of 15m/s

c) Express F4 in unit vector notation
http://img338.imageshack.us/img338/7817/forcefu9.th.jpg

Homework Equations


F=ma??

The Attempt at a Solution



a)

F1= 10N(cos 60i - sin 60j)= 5.00i- 8.66j

F2= 3N(cos 90i + sin 90j)= 3j

F3= 4N(-cos 20i- sin 20j)= -3.76i- 1.37j

b) F4= F1 + F2 + F3

F4= 5.00i - 8.66j + 3j - 3.76i - 1.37j= 1.24i - 7.03j

direction:
tan(theta)= y/x = -7.03/1.24= -5.6693
tan^-1(-5.6693)= -79.997

theta= -79.997 deg

magnitude:
F4= [tex]\sqrt{} (1.24)^2 + (-7.03)^2[/tex]= 7.14N

F4= 7.14N, -79.997 deg


I'm not sure how do I know if the magnitude and direction of the mass (strangely they now describe the mass as a particle) is maintaining a constant velocity of 15m/s?

c) for vector notation of F4 wouldn't it be

F4= 7.14N
theta= -79.997

x= F4 cos (theta)
y= F4 sin (theta)

x= 7.14 cos(-79.997) = 1.24i

y= 7.14 sin ( -79.997) = -7.03j

F4= 1.24i- 7.03j

~this is the same as before I converted it to cartesian coordinates...so Is this fine or is the previous part c have to do anything with the answer for d?
I didn't know what they wanted for b so I just converted the answer for c that I got without taking into account the 15m/s velocity requirement...


Do I need to use F= ma?? this wouldn't make sense b/c there is no place for velocity in the equation...


Help please..

Thank you =)
 
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Part a seems correct.

In order for the mass to move at constant velocity, it must have no net force acting on it, so F4 must act with equal magnitude, but opposite (direction) to the resultant (sum) of force F1 + F2 + F3.

F = ma is not necessary for this problem. For constant velocity a = 0.
 
In order for the mass to move at constant velocity, it must have no net force acting on it, so F4 must act with equal magnitude, but opposite (direction) to the resultant (sum) of force F1 + F2 + F3.

F = ma is not necessary for this problem. For constant velocity a = 0.

since F1 + F2 + F3= F4=
F4= 7.14N
theta= -79.997

but theta would equal -79.997 + 180= 100.003 deg

c.) I would then I guess just convert it to vector unit notation.

where
x= 7.14(cos 100.003)= -1.24i
y= 7.14(sin 100.003)= 7.03j

F4= -1.24i + 7.03j


I think this is fine right? Just making sure =D

Thanks Astronuc =)