contestada

Forces of 70 N at 130 degrees, and 20 N at an angle of 280 degrees, measured counter-clockwise from the positive x-axis, act on an object.
A. What are the components (F1x, F1y) of the first force force (in Newtons)?
B. What are the components (F2x, F2y) of the second force force (in Newtons)?
C. What are the components (Fx, Fy) of the resultant force (in Newtons)?
D. What is the magnitude of the resultant force (in Newtons)?
E. What is the angle of the resultant force with respect to x-axis?

Respuesta :

cjrodriguez89

Answer:

A. ) F₁ₓ = -45.0 N F₁y = 53.6 N

B.)  F₂ₓ = 3.48 N F₂y = -19.7 N

C.)  Fₓ = -41.5 N Fy = 33.9 N

D)  F = 53.6 N

E)  θ = -39. 2º (320.8º)

Explanation:

A)

  • Applying simple trig, like definitions of cos and sin of an angle, we can get the x- and y- components of F₁, as follows:

       [tex]F_{x1} = F_{1} * cos (130) = 70 N * cos (130) = -45 N (1)\\F_{y1} = F_{1} * sin (130) = 70 N * sin (130) = 53.6 N (2)[/tex]

B)

  • Repeating for F₂:

       [tex]F_{x2} = F_{2} * cos (280) = 20 N * cos (280) = 3.48 N (3)\\F_{y2} = F_{2} * sin (280) = 20 N * sin (280) = -19.7 N (4)[/tex]

C)  

  • The x- and y- components of the resultant force, are just the algebraic

        sum of the x- and - y components of F₁ and F₂:

  • Fₓ = Fₓ₁ + Fₓ₂ = -45 N + 3.48 N =  -41.5 N (5)
  • By the same token, Fy can be written as follows:
  • Fy = Fy₁ + Fy₂ = 53.6 N + (-19.7 N) = 33.9 N (6)

D)

  • The magnitude of the resultant force can be obtained applying the Pythagorean Theorem to Fx and Fy, as follows:

       [tex]F_{t} =\sqrt{F_{x} ^{2} + F_{y} ^{2} } = \sqrt{(-41.5N)^{2} +(33.9N)^{2}} = 53.6 N (7)[/tex]

E)

  • Finally the angle regarding the x- axis of the resultant force vector, can be obtained using the definition of the tangent of an angle, as follows:

       [tex]\theta = arc tg \frac{33.9N}{(-41.5N)} = arc tg (-0.817) = -39. 2 \deg[/tex]