# Two forces, F⃗ 1 and F⃗ 2, act at a point, as shown in the picture. (Figure 1) F⃗ 1 has a magnitude of 8.20N and is directed at an angle of α = 55.0∘ above the negative x axis in the second quadrant. F⃗ 2 has a magnitude of 6.40N and is directed at an angle of β = 53.4∘ below the negative x axis in the third quadrant.

a)
What is the x component Fx of the resultant force?
B)
What is the y component Fy of the resultant force?
C)
What is the magnitude F of the resultant force?
D)
What is the angle γ that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.

## A) Fx = F1x + F2x

To find the x component of F1, we can use:
Fx1 = F1 * cos(α)
Where F1 = 8.20N and α = 55.0°
So, Fx1 = 8.20N * cos(55.0°)

To find the x component of F2, we can use:
Fx2 = F2 * cos(180° - β)
Where F2 = 6.40N and β = 53.4°
So, Fx2 = 6.40N * cos(180° - 53.4°)

Fx = Fx1 + Fx2

B) Fy = F1y + F2y
To find the y component of F1, we can use:
Fy1 = F1 * sin(α)
Where F1 = 8.20N and α = 55.0°
So, Fy1 = 8.20N * sin(55.0°)

To find the y component of F2, we can use:
Fy2 = F2 * sin(180° - β)
Where F2 = 6.40N and β = 53.4°
So, Fy2 = 6.40N * sin(180° - 53.4°)

Fy = Fy1 + Fy2

C) To find the magnitude F of the resultant force, we can use the Pythagorean theorem:
F = √(Fx^2 + Fy^2)

D) To find the angle γ that the resultant force forms with the negative x axis, we can use the tangent function:
γ = tan^(-1)(Fy / Fx)

Now, let's calculate the values using the given information.

## To find the x and y components of the resultant force, we can use the following equations:

Fx = F1x + F2x
Fy = F1y + F2y

Where F1x and F1y are the x and y components of F1, and F2x and F2y are the x and y components of F2.

a) Finding Fx:

Fx = F1x + F2x
Fx = F1 * cos(α) + F2 * cos(β)

Given:
F1 = 8.20N
α = 55.0°
F2 = 6.40N
β = 53.4°

Using the given values:

Fx = 8.20N * cos(55.0°) + 6.40N * cos(53.4°)
Fx ≈ 5.83N

So the x component Fx of the resultant force is approximately 5.83N.

b) Finding Fy:

Fy = F1y + F2y
Fy = F1 * sin(α) + F2 * sin(β)

Using the given values:

Fy = 8.20N * sin(55.0°) + 6.40N * sin(53.4°)
Fy ≈ -5.94N

So the y component Fy of the resultant force is approximately -5.94N.

c) Finding the magnitude F of the resultant force:

F = √(Fx^2 + Fy^2)

Using the values previously calculated:

F = √(5.83N^2 + (-5.94N)^2)
F ≈ 8.27N

So the magnitude F of the resultant force is approximately 8.27N.

d) Finding the angle γ:

We can use the following equation to find the angle:

γ = arctan(Fy / Fx)

Using the values previously calculated:

γ = arctan((-5.94N) / 5.83N)
γ ≈ -46.8°

Since γ is measured clockwise from the negative x axis, the angle can be expressed as -46.8°.

## To find the x component Fx of the resultant force, we need to add the x components of F1 and F2. The x component can be found using the formula:

Fx = F * cos(θ)

where F is the magnitude of the force and θ is the angle it makes with the x-axis.

a) Given that F1 = 8.20 N and the angle α = 55.0°, we can find the x component F1x using:

F1x = F1 * cos(α)

Similarly, for F2 = 6.40 N and the angle β = 53.4°, we can find the x component F2x using:

F2x = F2 * cos(β)

Adding these x components together, we get the x component Fx of the resultant force.

Now, let's calculate the x component Fx.

1) For F1x:
F1x = 8.20 N * cos(55.0°)

Use a calculator to calculate the cosine of 55.0°: cos(55.0°) ≈ 0.5736

F1x ≈ 8.20 N * 0.5736 ≈ 4.70 N

2) For F2x:
F2x = 6.40 N * cos(β)

Using the given angle β = 53.4°, we have:
cos(β) ≈ cos(53.4°) ≈ -0.602

F2x ≈ 6.40 N * -0.602 ≈ -3.8528 N

Fx = F1x + F2x
≈ 4.70 N + (-3.8528 N)
≈ 0.8472 N

Therefore, the x component Fx of the resultant force is approximately 0.8472 N.

Next, let's find the y component Fy of the resultant force. The y component can be found using the formula:

Fy = F * sin(θ)

where F is the magnitude of the force and θ is the angle it makes with the x-axis.

b) Using the same approach, we can find the y components F1y and F2y.

1) For F1y:
F1y = F1 * sin(α)

F1y = 8.20 N * sin(55.0°)

Use a calculator to calculate the sine of 55.0°: sin(55.0°) ≈ 0.8192

F1y ≈ 8.20 N * 0.8192 ≈ 6.7134 N

2) For F2y:
F2y = F2 * sin(β)

Using the given angle β = 53.4°, we have:

sin(β) ≈ sin(53.4°) ≈ -0.7966

F2y ≈ 6.40 N * -0.7966 ≈ -5.1002 N

Fy = F1y + F2y
≈ 6.7134 N + (-5.1002 N)
≈ 1.6132 N

Therefore, the y component Fy of the resultant force is approximately 1.6132 N.

To find the magnitude F of the resultant force, we can use the Pythagorean theorem:

F = sqrt(Fx^2 + Fy^2)

c) Plugging in the values we found:
F = sqrt((0.8472 N)^2 + (1.6132 N)^2)

Use a calculator to evaluate the square root expression:

F ≈ sqrt(0.7175 N^2 + 2.6017 N^2) ≈ sqrt(3.3192 N^2) ≈ 1.8223 N

Therefore, the magnitude F of the resultant force is approximately 1.8223 N.

Now, let's find the angle γ that the resultant force forms with the negative x axis. The angle α + β gives the direction of the resultant force. In this problem, assume that positive angles are measured clockwise from the negative x-axis.

d) To find the angle γ, we can use the tangent formula:

tan(γ) = Fy / Fx

γ = atan(Fy / Fx)

Plugging in the values we found for Fy and Fx:

γ = atan(1.6132 N / 0.8472 N)

Use a calculator to calculate the arctangent (atan) of the ratio:
γ ≈ atan(1.9027)

γ ≈ 46.35°

Therefore, the angle γ that the resultant force forms with the negative x-axis is approximately 46.35°.

## F1 = 8.2N[125o]

F2 = 6.40N[233.4o]

A. Fx = 8.2*cos125 + 6.40*cos233.4 =
-8.519 N.

B. Fy = 8.2*sin125 + 6.40*sin233.4 =
1.579 N.

C. F^2 = -8.519^2 + 1.579^2 = 75.07
F = 8.66 N.

D. Tan Ar=Fy/Fx = 1.579/-8.519=-0.18535
Ar = -10.5o = Reference angle.

A = -10.5 + 180 = 169.5o = 10.5o Above
the negative x axis.