Question

The forces F1 = (2i + bj)N, F2 = (-i + 2j)N and F3 = (ai -4j)N act through the points with position vectors r1 = (i + 3j)m, r2 = (xi + 5j)m and r3 = (-i + j)m respectively.
Given that the system of forces is equivalent to a couple of magnitude 12Nm, find:
(a) The values of the constants a and b.
(b) The possible values of the contant x.

Answer 1

As the system of forces is equivalent to a couple of magnitude 12Nm, then

$\vec F_1+\vec F_2+\vec F_3=\vec 0$

(2i + bj) + (-i + 2j) + (ai -4j) = (a+1)i + (b-2)j = 0

a= -1 and b=2.

The magnitude of a couple is equal to

$\vec r_1\times\vec F_1 + \vec r_2\times\vec F_2 + \vec r_3\times\vec F_3 = (i + 3j)\times (2i + 2j) + (xi + 5j)\times(-i + 2j) +(-i + j)\times(-i -4j) = -4 + (2x+5) + 5 =2x+6$ 12Nm is an absolute value (unsigned) of magnitude, therefore we have to consider both signs:

2x+6 = 12 implies x=3

2x+6 =-12 implies x=-9

Answer. (a) a= -1, b=2; (b) x=3 or x=-9.