1. Feb 22, 2012

netrunnr

1. The problem statement, all variables and given/known data

There are two forces on the 3.2 kg box in the overhead view of the figure below but only one is shown. The figure also shows the acceleration of the box. F1 is in Q I and F2 is in Q III
a = 12m/s2

there is a picture here that I do not know how to upload -- it shows F1 in the +x direction ( 0º ) and F2 in Q III in the -x and -y direction at 30º from the y axis in the SW direction
then below that it gives the a=12m/s2

I am not sure if that means a=12m/s2 for F2 or if that is for the fnet = ma where that is a......
(a) Find the second force in unit-vector notation.
(b) Find the second force as a magnitude and direction.

2. Relevant equations and 3. The attempt at a solution

(a) Find the second force in unit-vector notation.

Fnet = ma
so Fnet = 3.2kg * 12m/s2 = 38.4N
F2 = 38.4N - F1 = 38N - 20N = 18.4

(b) Find the second force as a magnitude and direction.

F2=ma - F1 = [3.2kg * 12m/s2]-20N = 18.4N
F2x = 18 cos 240º = -9i
F2y = 18 sin 240º = -15.59i

I am supposed to be getting -39.2iand -33.3j for F2 for part A and 51.4N (-140)º for part b. if they gave me ø how am I supposed to get -140º for ø? I am very confused with this question.

2. Feb 22, 2012

PeterO

To get a picture of what is going on, try the following.

You have calculated that the net Force is 38.4 N, so draw a circle of Radius 38.4 on your axes [ you might need bigger axes, or don't draw the 20 Force so long ]

Since you are adding F1 and F2 - done by connecting the vectors head to tail, translate your F2 so that is begins at the end of F1.
It has to be long enough to reach a point on the circle.

That will show you what size F2 is, and hopefully your Pythagoras and trig skills are up to solving it.
You might even end up solving some simultaneous equations.

3. Feb 23, 2012

netrunnr

if I that I get the law of sins for 20N/ø =r/150º
I still have two unknowns here :(

4. Feb 23, 2012

PeterO

You could find the equation of the straight line that has F2 as part of it (referring to the x-y axes you have overlaying your vectors I think it will be something like y = √3x - 20√3
You know the equation of the circle is x2 + y2 = (38.4)2

Solve those two simultaneously and you have the end of the F2 vector. The resultant force is thus from (0,0) to there.

NOTE: being a quadratic equation there will be two points of intersection, but by looking at your diagram you will be able to recognise which one if the correct one.