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three forces equal to 3p,5p,7p act simultaneously along the three side AB, BC,and CA of equilateral triangle ABC of side a. find the magnitude, direction and position of the resultant.​

Answer :

onyebuchinnaji
  1. The resultant of the vectors is 3.46 p
  2. The direction of the vector is 29.98⁰
  3. The position of the vector is 150⁰

The given parameters include;

  • side AB = 3p
  • side BC = 5p
  • side CA = 7p

From the image uploaded;

The resolution of the vectors along the sides of the triangle is as follows;

  • the y-component = force x sin(θ)
  • the x-component = force x cos (θ)

forces -------angle (θ) --------y-component-------x-component

3p -------------  60⁰ --------- ----2.598 p ------------------- 1.5 p

5p -------------- 60⁰ ------------ (-4.33p) ------------------- 2.5 p

7p -------------- 60⁰ -------------  0   -------------------------- (-7p)

sum:                                   ( -1.732 p) y                  (-3P)x

The resultant of the vectors:

R² = (-1.732 p)²  +  (-3P)²

R² = 11.998 p²

R = √(11.998 p²)

R = 3.46 p

The direction of the vector:

[tex]\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{-1.732}{-3} )\\\\\theta = tan^{-1} (0.577)\\\\\theta = 29.98 \ ^0[/tex]

The position of the vector = 180 - θ = 180 - 29.98 = 150⁰ (second quadrant)

To learn more about resultant and direction of forces, please visit: https://brainly.in/question/21981854

${teks-lihat-gambar} onyebuchinnaji

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