Question

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 1

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:

$\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{-1.732}{-3} )\\\\\theta = tan^{-1} (0.577)\\\\\theta = 29.98 \ ^0$

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

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