# Question Video: Analysis of the Equilibrium of a Uniform Rod Placed Horizontally on Two Supports

A uniform rod 𝐴𝐵 having a length of 1.3 m and weighing 147 N is resting in a horizontal position on two supports, where the support 𝐶 is at the end 𝐴 and 𝐷 is at a distance 𝑥 from the end 𝐵. Find the reaction of the support 𝑅_(𝐶) and the distance 𝑥, given that 𝑅_(𝐶) = (2/5)𝑅_(𝐷).

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### Video Transcript

A uniform rod 𝐴𝐵 having a length of 1.3 meters and weighing 147 newtons is resting in a horizontal position on two supports, where the support 𝐶 is at the end 𝐴 and 𝐷 is at a distance 𝑥 from the end 𝐵. Find the reaction of the support 𝑅 𝐶 and the distance 𝑥, given that 𝑅 𝐶 is equal to two-fifths 𝑅 𝐷.

Because the rod is resting, we know that it is in equilibrium, which means two conditions are met. The net force on the rod is zero, and the net moment of force about any point on the rod is also zero. Moments of force are always calculated about some arbitrary reference point. Let’s call the magnitude of the force of interest 𝐹. And we’ll call the distance between where the force is acting and the reference point 𝑑. If the line connecting where the force is acting and the reference point is perpendicular to the line of action of the force itself, then the moment of this force about the reference point takes on a specially simple form.

The magnitude of the moment of force in this situation is given by the magnitude of the force times the distance to the reference point. We specify that this is the magnitude because moments of force also have a sign. If we imagine that this orange line were actually a solid rod pivoting about the reference point, the force as drawn would be pulling it in a clockwise direction. If the force pointed in the opposite direction, then it would be pulling the rod in a counterclockwise direction. The sign of the moment of force is a convention that we choose, which tells us whether the force is pointing clockwise or counterclockwise. Usually we arbitrarily pick the counterclockwise direction to be positive, but as long as we’re consistent, it really doesn’t matter.

Now that we have several equations and conditions, let’s draw a diagram to organize the information that we are given. Here we have a rod with a length of 1.3 meters resting in a horizontal position on the supports 𝐶 and 𝐷. The support 𝐶 is located at the end 𝐴 as specified in the statement, and 𝐷 is located somewhere in the middle a distance of 𝑥 from the end 𝐵. There are also several forces acting on the rod. The weight of the rod is 147 newtons. And because the rod is uniform, the weight acts at its midpoint. Each support also provides a reaction force pushing up on the rod to counteract its weight.

Using the variables defined in the question, we’ll call the reaction force at the support 𝐷 𝑅 𝐷. And we’ll call the reaction force at the support 𝐶 𝑅 𝐶, where 𝑅 𝐶 is two-fifths of 𝑅 𝐷. Finally, to complete this diagram, we’ll just include that the distance from point 𝐴 to the support 𝐷 is 1.3 meters minus 𝑥 and that the weight 147 newtons is acting half of 1.3 meters, which is 0.65 meters or 65 centimeters, from either end. All right, let’s now use our two equilibrium conditions to solve for the two quantities that we’re looking for, the force 𝑅 𝐶 and the distance 𝑥.

To make sure the net force is zero, we add all of the forces pointing in one direction and subtract all of the forces pointing in the opposite direction. This works out especially well here because all of the forces are parallel. So we have two-fifths times 𝑅 𝐷, which is just 𝑅 𝐶 expressed in terms of 𝑅 𝐷, plus 𝑅 𝐷 minus 147 equals zero. Two-fifths 𝑅 𝐷 plus 𝑅 𝐷 is seven-fifths 𝑅 𝐷. Adding 147 to both sides gives us seven-fifths 𝑅 𝐷 equals 147. We can solve for 𝑅 𝐷 by multiplying both sides by five-sevenths. We find that 𝑅 𝐷 is 105 newtons. Then either by using the formula given to us in the statement or the fact that 𝑅 𝐷 plus 𝑅 𝐶 is the weight of 147 newtons, we find that 𝑅 𝐶 is 42 newtons.

Now that we have found 𝑅 𝐶, one of the quantities that we’re looking for, we can use our condition for the net moment of force to find 𝑥, the other quantity that we’re looking for. Let’s choose our reference point to be the end of the rod at 𝐵. The reference point is on the rod. And all of the forces are acting on the rod. So the line connecting the reference point to where the forces are acting is just the rod itself. Since 𝑅 𝐷, the weight, and 𝑅 𝐶 are all perpendicular to the rod, we can use our equation for the magnitude of the moments, force times distance. For the signs of the moments, observe that 𝑅 𝐷 and 𝑅 𝐶 point in the same orientation, while the weight points in the opposite orientation. So whatever sign we choose, the moment from 𝑅 𝐷 and from 𝑅 𝐶 will have the same sign and the moment from the weight will have the opposite sign.

Let’s arbitrarily choose the sign for the moments of 𝑅 𝐷 and 𝑅 𝐶 to be positive. The distance from 𝑅 𝐷 to 𝐵 is 𝑥, so its moment is 105 newtons times 𝑥. The distance from 𝑅 𝐶 to 𝐵 is 1.3 meters, so its moment is 42 newtons times 1.3 meters. Finally, the distance from the weight of the rod to 𝐵 is the same as the distance from the weight of the rod to 𝐴, which is 0.65 meters. So the moment of the weight is negative 147 newtons times 0.65 meters about 𝐵.

And as with the net force, this whole expression must evaluate to zero. If we plug all of the numerical parts into a calculator, we get negative 40.95. If we add 40.95 to both sides, we get 105𝑥 equals 40.95, which we can then solve for 𝑥 by dividing both sides by 105. This gives us that 𝑥 equals 0.39 meters, or equivalently 𝑥 is 39 centimeters. And with that, we’ve found both 𝑅 𝐶 and 𝑥, the true quantities that we were looking for.