Video Transcript
𝐴𝐵 is a uniform rod having a length of 78 centimeters and weighing 155 newtons. The rod is resting horizontally on two supports, 𝐴 and 𝐶, where 𝐶 is 13 centimeters away from 𝐵. Determine the minimum weight 𝑤 to be suspended at 𝐵 so that there is no pressure at 𝐴, and find out the pressure on 𝐶 at that instant.
The fact that the rod is resting means that it is in equilibrium, which means two conditions are met for the forces acting on the rod. These two conditions are, first, that the net force or the sum of all the forces acting on the rod is zero. And the second condition is that the net moment of force, that is, the sum of the moments of each force acting on the rod is also zero about any given point. The moment of the force about a reference point is the magnitude of the force times the distance to the reference point, as long as the line of action of the force is perpendicular to the line to the reference point. Okay, so with these conditions, let’s draw a diagram to organize the information we have about which forces are acting on the rod and where they’re acting.
Here’s our rod with the two supports 𝐴 and 𝐶. The length of the rod is 78 centimeters and the distance from 𝐶 to 𝐵 is 13 centimeters. There are also four forces acting on the rod. There is the weight of the rod, 155 newtons, which acts directly in the center pointing downward because 𝐴𝐵 is uniform. Being in the center, this weight is acting 39 centimeters from either end. There is also the additional weight 𝑤, which is attached at point 𝐵. The other two forces acting on the rod are the reaction forces from the two supports.
The combined force of the weight of the rod and the weight at 𝐵 place pressure on the support at 𝐴 and the support at 𝐶. Newton’s third law guarantees that as the rod exerts pressure force on these two supports, these supports exert a force of reaction onto the rod with equal magnitude but opposite direction. We’ll call these two forces 𝑅 sub 𝐴 and 𝑅 sub 𝐶 for the reaction at 𝐴 and the reaction at 𝐶. There are actually two important things that we already know about these two forces. First, we’re looking for the condition where there is no pressure at 𝐴. But if there is no pressure force, there was also no reaction force. So we’re looking for the condition where 𝑅 sub 𝐴 is zero.
Secondly, because the magnitude of the pressure on the support 𝐶 and the magnitude of the reaction 𝑅 sub 𝐶 are the same, if we can determine 𝑅 sub 𝐶, we will determine the pressure that we’re looking for. All right, now that we have these forces and distances, let’s apply our conditions for equilibrium to solve for what we’re looking for. Let’s start with the condition that the moments of force sum to zero. We’re starting with this condition because since we can express a moment of force as force times distance, if the distance from the force to the reference point is zero — that is, the force is acting at the reference point — it contributes no moment.
Since this is true regardless of the magnitude of the force, if we choose our reference point to be the point 𝐵 or where the support meets the rod at point 𝐶, we will either eliminate the unknown weight or the unknown reaction from our calculation. By eliminating one unknown from the calculation, we’ll be able to directly solve for the other unknown. Note that since there is no reaction force at point 𝐴, it also provides no moment because the magnitude of the force is zero.
All right, let’s start with our reference point at point 𝐵. The forces we need to work with are then 𝑅 𝐶 acting 13 centimeters away and the weight of the rod acting 39 centimeters away. Furthermore, both of these forces are perpendicular to the rod, which is along the line connecting the force to the reference point, so we can use force times distance to calculate the appropriate moments. Since the reaction force points up and the weight points down and both are to the left of the reference point, this means they have opposite orientations relative to the reference point. So their moments have opposite signs.
Let’s arbitrarily choose the moment of 𝑅 𝐶 about 𝐵 to be positive. Then we have 𝑅 𝐶 times 13, the moment of 𝑅 𝐶 about 𝐵, minus 155 times 39, the moment of the weight of the rod about 𝐵, equals zero. If we divide both sides by 13, we get 𝑅 𝐶 minus 155 times three equals zero because 39 divided by 13 is three and zero divided by 13 is still zero. Adding 155 times three to both sides, we find that 𝑅 𝐶 is 155 times three or 465 newtons. But as we already saw, this is exactly the pressure that we’re looking for. So half of our answer is that the pressure on the support 𝐶 when the rod is balanced as described is 465 newtons.
Now all that’s left is to find the weight. We could perform exactly the same type of calculation, but this time with the reference point where the support 𝐶 meets the rod. Alternatively, now that we know a value for 𝑅 𝐶, we can use our condition for the sum of the forces to solve for the only remaining unknown, which is the weight 𝑤. Whichever method we pick, we’re still relying on the fact that the magnitude of the force at 𝐴 is zero, so it provides no moment and no contribution to the net force. Let’s use the condition for the sum of the forces, and we’ll clear away our previous calculation to make room for this new one.
Since all of the forces that we’re interested in are parallel, we add up all of the forces pointing in one direction and subtract all the forces pointing in the opposite direction. This gives us 465, the known value for 𝑅 𝐶, minus 155 minus 𝑤 equals zero. 465 minus 155 is 310. And adding 𝑤 to both sides gives us that 𝑤 is 310, which means the weight acting at 𝐵 needed to balance the rod as described is 310 newtons.
