In the figure below, 𝐅 sub one
and 𝐅 sub two are two parallel forces measured in newtons, where 𝑅 is their
resultant. If 𝑅 is equal to 30 newtons,
𝐴𝐵 equals 36 centimeters, and 𝐵𝐶 equals 24 centimeters, determine the
magnitude of 𝐅 sub one and 𝐅 sub two.
In this question, we have two
parallel coplanar forces, 𝐅 sub one and 𝐅 sub two, acting in opposite
directions. We are also given the resultant
force 𝑅, which is equal to 30 newtons. The distance from point 𝐴 to
𝐵 is 36 centimeters, and the distance from 𝐵 to 𝐶 is 24 centimeters. The lines of action of our
forces 𝐅 sub one, 𝐅 sub two, and 𝑅 are not perpendicular to the line segment
𝐴𝐶. However, as our three forces
are parallel, we can add the angle 𝜃 to our diagram as shown.
We can calculate the
perpendicular components of these forces using our knowledge of right angle
trigonometry. These are equal to 𝐅 sub one
sin 𝜃, 𝐅 sub two sin 𝜃, and 𝑅 sin 𝜃. These will be useful when we
come to take moments, as the moment of a force is equal to the magnitude of
force multiplied by the perpendicular distance to the point at which we are
taking moments. Whilst we can take moments
about any point on our line, in this question, we will do so about point 𝐴.
We will consider moments acting
in the counterclockwise direction to be positive and those acting in the
clockwise direction to be negative. This means that the moment 𝑀
sub one of the force 𝐅 sub one acts in the positive direction and is equal to
𝐅 sub one sin 𝜃 multiplied by 36. We can repeat this process for
𝑀 sub two, which is the moment of the force 𝐅 sub two. As this acts in the negative
direction, this is equal to negative 𝐅 sub two sin 𝜃 multiplied by 60.
Our expressions for 𝑀 sub one
and 𝑀 sub two can be simplified as shown. We know that the distance 𝑥
from the line of action of the resultant force to the point at which we are
taking moments is equal to the sum of the moments divided by 𝑅. In this case, we are taking
moments about the point where the resultant acts. Therefore, 𝑥 is equal to
zero. This means that the sum of our
two moments must equal zero. 36 multiplied by 𝐅 sub one sin
𝜃 plus negative 60 multiplied by 𝐅 sub two sin 𝜃 equals zero. Since sin 𝜃 cannot be equal to
zero, we can divide through by this. We can also divide through by
12 such that three 𝐅 sub one minus five 𝐅 sub two equals zero. As there are two unknowns here,
we will call this equation one.
We will now consider the
resultant force and the fact that this is equal to the sum of the other
forces. Going back to our initial
diagram, if we let the positive direction be vertically upwards, we have 𝑅 is
equal to 𝐅 sub one plus negative 𝐅 sub two. Since 𝑅 is equal to 30
newtons, we have 30 is equal to 𝐅 sub one minus 𝐅 sub two. Adding 𝐅 sub two to both sides
of this equation, we have 𝐅 sub one is equal to 30 plus 𝐅 sub two. We will call this equation
two. And we now have a pair of
simultaneous equations that we can solve by substitution.
One way of doing this is to
substitute the expression for 𝐅 sub one in equation two into equation one. This gives us three multiplied
by 30 plus 𝐅 sub two minus five 𝐅 sub two is equal to zero. Distributing the parentheses
gives us 90 plus three 𝐅 sub two. The left-hand side then
simplifies to 90 minus two 𝐅 sub two. By adding two 𝐅 sub two to
both sides and then dividing through by two, we have 𝐅 sub two is equal to
45. Substituting this value back
into equation two gives us 𝐅 sub one is equal to 30 plus 45, which is equal to
75. The magnitude of the two forces
𝐅 sub one and 𝐅 sub two are 75 newtons and 45 newtons, respectively.