Appliced mathematics (Vectors

F1, F2 and F3 are in equilibrium,

(5i + 6j) + (ai - 4j) + (-6i + bj) = 0

(a - 1)i + (2 + b)j = 0

a = 1, b = -2

Line of actions:

L1: ci + j + t(5i + 6j)

L2: 2i - j + m(i - 4j)

L3: 3i + 2j + n(-6i - 2j)

We have,

c + 5t = 2 + m = 3 - 6n

1 + 6t = -1 - 4m = 2 - 2n

Solving, c = 1, m = -8/13, n = 7/26, t = 1/13

Putting into L2,

Point of intersection: 18/13 i + 19/13 j



b. F3 = 6i + 2j

Resultant force of the system, R

= F1 + F2 + F3

= (5i + 6j) + (i - 4j) + (6i + 2j)

= 12i + 4j

Moment by F1 , F2 , F3

=
│i j k│ │i j k│ │i j k│
│1 1 0│ + │2 -1 0│ +│3 2 0│
│5 6 0│ │1 -4 0│ │6 2 0│

= -12k (clockwise)

Adding the couple, the magnitude of the moment = 39 units

Let @i + #j be the position where the resultant acts through

│i j k│
│@ # 0│ = 39k
│12 4 0│

4@ - 12# = 39

Set # = 3, @ = 3/4

So, the vector equation of its line of action:

3/4 i + 3j + t(12i + 4j)

where t is any real number



2. F4 = -(F1 + F2 + F3)

= -[(3i - j + 2k) + (-i - 4j + k) + (i + j - 2k)]

= -3i + 4j - k

Moment by F1, F2 and F3

=

│i j k│ │i j k│ │i j k│

│3 0 -1│ + │2 -4 0│ +│0 -3 5│

│3 -1 2│ │-1 -4 1│ │1 1 -2│

= -4i - 12k

Let ai + bj + ck be the position where F4 acts through

│i j k│


│a b c│ = -4i - 12k

│-3 4 -1│

(-b - 4c)i + (a - 3c)j + (4a + 3b)k = -4i - 12k

Set z = t, y = -4 - 4t, x = 3t, t is any real number

Put z = 0, y = -4, x = 0

So, the line of action: -4j + s(-3i + 4j - k), s is any real number