Vector Addition Homework: Find F2 Magnitude & Direction

[netrunnr]

Homework Statement

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 f_net = ma where that is a...
(a) Find the second force in unit-vector notation.
(b) Find the second force as a magnitude and direction.

Homework Equations

and

The Attempt at a Solution

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

F_net = ma
so F_net = 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
F_2x = 18 cos 240º = -9_i
F_2y = 18 sin 240º = -15.59_i
I am supposed to be getting -39.2_iand -33.3_j 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.

 

[PeterO]

Homework Statement

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 f_net = ma where that is a...
(a) Find the second force in unit-vector notation.
(b) Find the second force as a magnitude and direction.

Homework Equations

and

The Attempt at a Solution

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

F_net = ma
so F_net = 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
F_2x = 18 cos 240º = -9_i
F_2y = 18 sin 240º = -15.59_i



I am supposed to be getting -39.2_iand -33.3_j 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.

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.

 

[netrunnr]

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

 

[PeterO]

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

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.