Video: Equilibrium of a Nonuniform Body Resting on a Horizontal Surface

𝐴𝐡 is a rod having a length 120 cm and weighing 12 N, which is acting at a point 15 cm away from 𝐴. Given that the rod is resting on a support at its midpoint, determine the reaction of the support 𝑅, and find the weight π‘Š that should be suspended from the end 𝐡 to make the rod in equilibrium in a horizontal position.

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Video Transcript

𝐴𝐡 is a rod having a length 120 centimeters and weighing 12 newtons, which is acting at a point 15 centimeters away from 𝐴. Given that the rod is resting on a support at its midpoint, determine the reaction of the support 𝑅 and find the weight π‘Š that should be suspended from the end 𝐡 to make the rod in equilibrium in a horizontal position.

The forces acting on an object in equilibrium satisfy two conditions. The first condition is that the sum of all the forces, that is, the net force acting on the object, is zero. And the second is that the net moment of force, that is, the sum of all the individual moments, is also zero. The moment of a force about some reference point is given by the magnitude of the force times the distance between the point where the force is acting and the reference point. In particular, we use this formula when the line of action of the force is perpendicular to the line connecting where the force is acting and the reference point.

Furthermore, such forces have either a clockwise or counterclockwise orientation about the reference point. When we find the net moment of force, we add all of the moments from the forces with one orientation and subtract all of the moments from the forces with the opposite orientation. The choice of which orientation corresponds to positive moments and which orientation corresponds to negative moments is arbitrary as long as we are consistent.

Okay, now that we have the equations that we’ll need, let’s draw a diagram to organize the information that we have about the forces and where they are acting. Here we’ve drawn a rod 𝐴𝐡 with its length of 120 centimeters and the support located at its midpoint, which is 60 centimeters from either side. The forces acting on the rod are its weight of 12 newtons, which is acting 15 centimeters away from 𝐴. It’s interesting to note that because the weight is not acting at the midpoint of the rod, we know that the rod is not uniform even though we’re not explicitly told so.

The other forces acting on the rod are the reaction of its support, which we’ll call 𝐹 sub 𝑅, which acts vertically at the midpoint of the rod. This reaction force is one of the quantities that we’re looking for. Finally, there is also the unknown weight π‘Š acting at the point 𝐡, which is the other quantity that we’re looking for. Now that we have our information organized, let’s use our conditions for equilibrium to determine values for the unknown quantities.

We have two unknowns, 𝐹 sub 𝑅 and π‘Š, and two equations that we can set up, which should be enough to solve for the two unknowns. We can actually set these up in a clever way to directly find one of the two unknowns immediately. We note that if a force is acting at the reference point for the moment of force, then the distance in the moment of force formula is zero. In other words, the moment of any force acting at the reference point is zero. So if we cleverly choose a reference point as either the midpoint of the rod or the end of the rod 𝐡, then the unknown 𝐹 sub 𝑅 or the unknown π‘Š will have a zero moment about the reference point. So we’ll be able to solve directly for the other one.

To see this concretely, we’ll choose a reference point for the moment of force as the midpoint of the rod. Then, the moment of 𝐹 sub 𝑅 about the reference point is zero, and the only two forces that contribute nonzero moments are the 12-newton weight of the rod and the additional weight π‘Š at 𝐡. Let’s arbitrarily choose the moment of π‘Š to be positive. π‘Š is 60 centimeters from the reference point, so its moment is positive π‘Š times 60 centimeters. The 12-newton weight is 15 centimeters from 𝐴, and 𝐴 is 60 centimeters from the midpoint. So the 12-newton weight is 60 minus 15 or 45 centimeters from the midpoint.

Now, since the 12-newton weight and the weight at 𝐡 are pointing in the same direction but are located on opposite sides of the reference point, they have opposite orientations about the reference point. So, since the moment of π‘Š is positive, the moment of the 12-newton weight must be negative, and it will be negative 12 newtons times 45 centimeters. So this expression gives us the net moment of force about the midpoint of the rod, and we know from the condition of equilibrium that this net moment is zero.

Now we have an equation with a single unknown, so let’s solve for π‘Š. We first add 12 times 45 to both sides, and then we can observe that 60 is five times 12 and 45 is five times nine. Noting that we have five times 12 on both sides and the only factor left on the left-hand side is π‘Š and the only factor left on the right-hand side is nine, we conclude that π‘Š is nine newtons.

To find 𝐹 sub 𝑅, we can either use our condition for the moments of force a second time or we can now use our condition for the sum of the forces because π‘Š is known. To find the sum of the forces, we add all the forces pointing in one direction and subtract all the forces pointing in the opposite direction. So we have 𝐹 sub 𝑅 minus 12 minus nine equals zero. Adding 12 and nine to both sides and recalling that 12 plus nine is 21, we have that the reaction force from the support is 21 newtons.

So we have found that a weight of nine newtons acting at 𝐡 will keep the rod in equilibrium in a horizontal position. And the resulting reaction force from the support at the midpoint will be 21 newtons.

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