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.

 

[asteriatic]

I would like to provide some clarification and guidance on how to approach this problem.

Firstly, the given information is incomplete as it does not specify the magnitudes of the forces F1 and F2. Therefore, it is not possible to accurately solve for F2 using the given information.

Secondly, the acceleration of the box (a=12m/s2) is not relevant to finding the magnitude and direction of F2. It is only used to determine the net force acting on the box, which is not necessary for solving this problem.

Thirdly, the unit-vector notation for F2 would be [F2x, F2y] instead of just F2.

To solve for F2, we need to use vector addition and trigonometry.

(a) To find F2 in unit-vector notation, we need to determine the x and y components of F2.

F2x = F2 cos 30º (since the angle between F2 and the x-axis is 30º)
F2y = -F2 sin 30º (since the angle between F2 and the y-axis is 30º in the SW direction, which is equivalent to -30º)

(b) To find the magnitude and direction of F2, we can use the Pythagorean theorem and inverse trigonometric functions.

Magnitude:
|F2| = √(F2x² + F2y²)

Direction:
F2θ = tan⁻¹(F2y/F2x)

Note: The direction, F2θ, will be in the range of -90º to 90º, so we need to adjust for the actual direction of F2 by considering the quadrant in which it lies. Since F2 is in QIII (negative x and y), the actual direction of F2 will be in the range of -180º to -90º.

In conclusion, without knowing the magnitudes of F1 and F2, it is not possible to accurately solve for F2. Additionally, the given information does not specify the direction of F1, which is needed to solve for F2. I would recommend double-checking the given information and seeking clarification from the source before attempting to solve this problem.