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Math  /  Geometry

QuestionFind the resultant of the two forces using: a) graphical method b) Cosine and sine rule methods c) Force components.

Studdy Solution

STEP 1

1. Forces are vectors and can be added using vector addition.
2. The angles given are measured from the horizontal axis.
3. The graphical method involves drawing vectors to scale and finding the resultant.
4. The cosine and sine rules are applicable for non-right triangles.
5. Force components can be resolved into horizontal and vertical components.

STEP 2

1. Graphical Method
2. Cosine and Sine Rule Methods
3. Force Components Method

STEP 3

Graphical Method: Draw the vectors to scale on graph paper. Use a protractor to ensure the angles are accurate. Place the tail of the second vector at the head of the first vector. The resultant vector is drawn from the tail of the first vector to the head of the second vector.

STEP 4

Cosine and Sine Rule Methods: First, find the angle between the two forces. Since both angles are measured from the horizontal, the angle between the forces is:
θ=75∘−20∘=55∘ \theta = 75^\circ - 20^\circ = 55^\circ

STEP 5

Use the cosine rule to find the magnitude of the resultant force R R :
R2=F12+F22−2⋅F1⋅F2⋅cos⁡(θ) R^2 = F_1^2 + F_2^2 - 2 \cdot F_1 \cdot F_2 \cdot \cos(\theta)
Where F1=30 N F_1 = 30 \, \text{N} , F2=50 N F_2 = 50 \, \text{N} , and θ=55∘ \theta = 55^\circ .

STEP 6

Calculate:
R2=302+502−2⋅30⋅50⋅cos⁡(55∘) R^2 = 30^2 + 50^2 - 2 \cdot 30 \cdot 50 \cdot \cos(55^\circ)
R2=900+2500−3000⋅cos⁡(55∘) R^2 = 900 + 2500 - 3000 \cdot \cos(55^\circ)
R=3400−3000⋅cos⁡(55∘) R = \sqrt{3400 - 3000 \cdot \cos(55^\circ)}

STEP 7

Use the sine rule to find the direction of the resultant force:
sin⁡(α)F2=sin⁡(θ)R \frac{\sin(\alpha)}{F_2} = \frac{\sin(\theta)}{R}
Where α \alpha is the angle opposite F2 F_2 .

STEP 8

Force Components Method: Resolve each force into horizontal and vertical components.
For F1=30 N F_1 = 30 \, \text{N} at 75∘ 75^\circ :
F1x=30⋅cos⁡(75∘) F_{1x} = 30 \cdot \cos(75^\circ) F1y=30⋅sin⁡(75∘) F_{1y} = 30 \cdot \sin(75^\circ)
For F2=50 N F_2 = 50 \, \text{N} at 20∘ 20^\circ :
F2x=50⋅cos⁡(20∘) F_{2x} = 50 \cdot \cos(20^\circ) F2y=50⋅sin⁡(20∘) F_{2y} = 50 \cdot \sin(20^\circ)

STEP 9

Sum the components to find the resultant components:
Rx=F1x+F2x R_x = F_{1x} + F_{2x} Ry=F1y+F2y R_y = F_{1y} + F_{2y}

STEP 10

Calculate the magnitude of the resultant force:
R=Rx2+Ry2 R = \sqrt{R_x^2 + R_y^2}

STEP 11

Calculate the direction of the resultant force:
tan⁡(ϕ)=RyRx \tan(\phi) = \frac{R_y}{R_x}
Where ϕ \phi is the angle of the resultant force from the horizontal.
The resultant force and its direction can be calculated using any of the methods described. The exact numerical solution requires computation of the trigonometric functions and arithmetic operations.

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