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Question Video: Finding the Point at Which a Weight Is Added to a Horizontal Rod Resting in Equilibrium on Two Supports Mathematics

Jennifer lay on a horizontal uniform wooden plank of length 2.6 m and weight 16 kg-wt that was fixed at each end on two supports A and B. Given that the reactions of the two supports 𝑅_A and 𝑅_B are 68 kg-wt and 52 kg-wt, respectively, determine the distance between the point action of her weight and support A.

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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, 𝑅 sub A and 𝑅 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.

Let’s 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 𝑅 sub A and 𝑅 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 π‘₯ 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 𝐹 multiplied by 𝑑 and where 𝐹 is the force acting at a point and 𝑑 is the perpendicular distance from this point to the point at which we are taking moments.

Before we can calculate the distance π‘₯, we will calculate Jennifer’s weight force 𝑀. 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 𝑀 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 𝑀 minus 16 equals zero. 68 plus 52 minus 16 is equal to 104. And adding 𝑀 to both sides of our equation, we have 𝑀 is equal to 104 kilogram-weight. We can then add this force onto our diagram.

Next, we’ll 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. Jennifer’s weight force acts in a clockwise direction about point A. This means it will have a negative moment equal to negative 104 multiplied by π‘₯. 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 𝑅 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 π‘₯ plus negative 16 multiplied by 1.3 plus 52 multiplied by 2.6 is equal to zero. This simplifies to negative 104π‘₯ minus 20.8 plus 135.2 equals zero. Adding 104π‘₯ to both sides, we have 114.4 is equal to 104π‘₯. We can then divide through by 104 such that π‘₯ is equal to 11 over 10 or 1.1. The distance between the point of action of Jennifer’s weight and support A is 11 over 10 or 1.1 meters.

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