Question

Two forces, F⃗ 1 and F⃗ 2, act at a point, as shown in the picture. (Figure 1) F⃗ 1 has a magnitude of 9.80 N and is directed at an angle of α = 58.0 ∘ above the negative x axis in the second quadrant. F⃗ 2 has a magnitude of 5.20 N and is directed at an angle of β = 52.4 ∘ below the negative x axis in the third quadrant.

Part A:What is the x component Fx of the resultant force?

Part B:What is the y component Fy of the resultant force?

Part C:What is the magnitude F of the resultant force?

Part 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.

Answer 1

Answer:

Part A: Fx= -1.66 N

Part B: Fy= 5.25 N

Part C: F=5.51 N

Part D: $\gamma =72.41$

Explanation:

1. Relevant Data

$F1=9.8 N\\\alpha=58\\F2=5.2 N\\\beta= 52.4$

2. Calculate x component of F1, F2 and F

$F1x=9.8 N cos 58 = 1.17 N\\F2x=5.2 N cos 52.4=-2.83 N\\Fx=F1x+F2x\\Fx=1.17 N-2.83 N=-1.66 N$

3. Calculate y component of F1, F2 and F

$F1y=9.8 N sin 58 = 9.73 N\\F2y=5.2 N sin 52.4=4.48 N\\Fx=F1x-F2x\\Fx=9.73 N-4.47 N=5.25 N$

4. Calculate the magnitud of F

$F^2=Fx^2+Fy^2$

$F=\sqrt{Fx^2+Fy^2} \\F=\sqrt{(-1.66)^2+(5.25)^2}\\F= 5.51 N$

5. Calculate the angle with the negative x axis

The angle γ could be calculated using the components x and y of the resultant force:

$tan \gamma =\frac{Fy}{Fx} \\\gamma =tan^-1(\frac{Fy}{Fx} )\\\gamma =tan^-1(\frac{5.25}{-1.66} )\\\\\gamma=-72.41$

Considering that positive angles are measured clockwise from the negative x axis, then $\gamma=72,41$

Answer 2

The resolution of forces, allows the representation of a given force by two forces that point in the x and y-directions respectively

The correct values for the question are;

Part A: The x-component of the resultant, Fₓ is approximately -8.4·i

Part B: The y-component of the resultant $F_y$ is approximately ≈ 4.19 j

Part C: magnitude of the resultant force, F ≈ 9.4 N

Part D: The direction of the resultant force above the negative direction is approximately 24.0°

The reason the above values are correct is as follows:

The given parameters of the forces, F₁ and F₂ are;

The magnitude of the force F₁ = 9.80 N

The angle at which F₁ is directed, α = 58.0° above the negative x-axis

The magnitude of the force F₂ = 5.20 N

The angle at which F₁ is directed, β = 52.4° above the negative x-axis

Solution:

The x any y-component of the force, F₁, are given as force as follows;

Fₓ₁ = -9.80 × cos(58) i

$F_{y1}$ = 9.80 × sin(58) j

The x any y-component of the force, F₂, are given as force as follows;

Fₓ₂ = -5.20 × cos(52.4°) i

$F_{y1}$ = -5.20 × sin(52.4°) j

Part A: The x-component of the resultant, Fₓ is given by the sum of the x-components of each of the forces that make up the the resultant force

∴ Fₓ = -9.80 × cos(58) i + -5.20 × cos(52.4°) i ≈ -8.4·i

The x-component of the resultant, Fₓ ≈ -8.4·i

Part B: The y-component of the resultant, $F_y$ is given by the sum of the y-components of each of the forces that make up the the resultant force as follows;

$F_y$ = 9.80 × sin(58) j + (-5.20 × sin(52.4°)) j ≈ 4.19 j

$F_y$ ≈ 4.19 j

Part C: The magnitude of the resultant force, F is given as follows;

F = √((-8.4)² + (4.19)²) ≈ 9.4

The magnitude of the resultant force, F ≈ 9.4 N

Part D: The direction of the resultant force, θ, above the negative x-axis direction is given as follows;

$\theta = arctan \left(\dfrac{F_y}{F_x} \right)$

Therefore;

$\theta \ approximates \ to \arctan \left(\dfrac{4.19}{9.4} \right) \approx 24.0^{\circ}$

The direction of the resultant force above the negative x-axis direction, θ is approximately 24.0°

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