Video Transcript
Jennifer lay on a horizontal
uniform wooden plank of length 2.6 meters and weight 16 kilogram-weight that was
fixed at each end on two supports A and B. Given that the reactions of the two
supports, ๐
sub A and ๐
sub B, are 68 kilogram-weight and 52 kilogram-weight,
respectively, determine the distance between the point of action of her weight and
support A.
Letโs begin by sketching a diagram
to model the situation. We have a horizontal wooden plank
of length 2.6 meters fixed at two supports at each end, A and B. Since the plank is uniform, the
weight acts at the midpoint of the plank. In this question, we have a weight
of 16 kilogram-weight acting 1.3 meters from A. We have reaction forces at A and B
denoted ๐
sub A and ๐
sub B. And we are told these are equal to
68 kilogram-weight and 52 kilogram-weight, respectively. We are told that these forces occur
when Jennifer lies at some point on the plank. We will call the distance that we
need to calculate between the point of action of her weight and support A ๐ฅ
meters.
We can assume that the plank is in
equilibrium. And as such, the sum of the forces
acting on it equal zero. When a body is in equilibrium, the
sum of the moments also equal zero, where the moment of a force is equal to ๐น
multiplied by ๐ and where ๐น is the force acting at a point and ๐ is the
perpendicular distance from this point to the point at which we are taking
moments.
Before we can calculate the
distance ๐ฅ, we will calculate Jenniferโs weight force ๐ค. If we let the positive direction be
vertically upwards, we have forces acting in this direction of 68 and 52
kilogram-weight. Acting in the opposite direction,
we have ๐ค and the 16-kilogram-weight force. This means that these will be
negative. And since the sum of the forces
equal zero, we have 68 plus 52 minus ๐ค minus 16 equals zero. 68 plus 52 minus 16 is equal to
104. And adding ๐ค to both sides of our
equation, we have ๐ค is equal to 104 kilogram-weight. We can then add this force onto our
diagram.
Next, weโll take moments about some
point on the plank. Whilst we could do this at any
point, as we are measuring the distance from A, we will take moments about point A
and let positive moment be in the counterclockwise direction. Jenniferโs weight force acts in a
clockwise direction about point A. This means it will have a negative
moment equal to negative 104 multiplied by ๐ฅ. The weight of the plank also acts
in a clockwise direction about point A. This has a moment of negative 16
multiplied by 1.3, as it is a distance of 1.3 meters from A. Next, we have the reaction force of
point B. This acts in a counterclockwise
direction about point A and will therefore have a positive moment equal to 52
multiplied by 2.6.
It is important to note that the
reaction force ๐
sub A will have a moment equal to zero. This is because the force is acting
at the point we are taking moments, and the distance will therefore equal zero. We now have the equation negative
104 multiplied by ๐ฅ plus negative 16 multiplied by 1.3 plus 52 multiplied by 2.6 is
equal to zero. This simplifies to negative 104๐ฅ
minus 20.8 plus 135.2 equals zero. Adding 104๐ฅ to both sides, we have
114.4 is equal to 104๐ฅ. We can then divide through by 104
such that ๐ฅ is equal to 11 over 10 or 1.1. The distance between the point of
action of Jenniferโs weight and support A is 11 over 10 or 1.1 meters.