Line of action of the resultant

Question

Total left direction moments of a coplanar system of forces are given by G, 2G and 4G about the points (0,0) , (1,2) and (2,5) respectively. Find the magnitude and the equation of the line of action of the resultant. (G doesn't equal 0)

Answer 1

Given:

(x₀,y₀) = (0, 0) with Moment M₀ = G

(x₁,y₀) = (1, 2) with Moment M₁ = 2G

(x₂,y₂) = (2, 5) with Moment M₂ = 4G

Since M=F_R•d, we set up the moment equations:

M₀ ​= G = F_R​⋅d₀​,

M₁ ​= 2G = F_R​•d1​,

M₂​ = 4G = F_R​•d_2​.

We can now relate d₀ to d₁ and d₂:

Dividing M₁ by M₀:

2G = F_R•d₀ => d₁ = 2d₀

G F_R•d₁

Dividing M₂ by M₀:

4G = F_R•d₀ => d₂ = 4d₀

G F_R•d₂

Now, using the perpendicular distance formula for a point (x,y) from a general line Ax+By+C=0,

d = |Ax + By + C|

√A²+B²

and using the given points (0,0), (1,2), and (2,5), set up the three equations:

1. d₀ = |A(0) + B(0) + C| = |C|

√A² + B² √A² + B²

1. d₁ = |A(1) + B(2) + C| = |A + 2B + C|

√A² + B² √A² + B²

1. d₂ = |A(2) + B(5) + C| = |2A + 5B + C|

√A² + B² √A² + B²

Since d₁=2d₀​ and d₂=4d₀​, divide the equations:

d₀ = |C| = 1 => C = A + 2B

d₁ |A + 2B + C| 2

d₀ = |C| = 1 => C = 2A + 5B

d₁ |2A + 5B + C| 4 3 3

d₁ = |A + 2B + C| = 1 => |2A + 4B + 2C| = |2A + 5B + C|

d₂ |2A + 5B + C| 2

Cancel A in the third equation:

|2A + 4B + 2C| = |2A + 5B + C|

C = B

Plugging into C = A + 2B:

C = A + 2C

C = -A

Thus, A = −C and B = C. The equation of the resultant line is:

(−C)x + (C)y + C = 0

C(y - x + 1) = 0

Dividing both sides by C, since C ≠ 1:

y = x - 1

For the resultant force, express d₀​ using the perpendicular distance formula:

d₀ = |C| = |C| = |C| = |C| = 1

√A² + B² √(-C)² + (C)² √(2C²) C√2 √2

Since M₀ = F_R•d₀:

G = F_R(1/√2)

F_R = G√2

The final results are:

Resultant Force: F_R = G√2

Resultant Line Equation: y = x - 1