A uniform seesaw is balanced on a fulcrum located 3.0 meters from the left end as shown. The smaller boy on the right has a mass of 40 kilograms, and the bigger boy on the left has a mass of 80 kilograms. What is the mass of the board?
The problem statement tells us that the boy on the right has a mass of 40 kilograms; we’ll call that 𝑚 sub 𝑟. The boy on the left we’re told has a mass of 80 kg; we’ll call that 𝑚 sub 𝐿. We want to know the mass of the board that the boys are sitting on; we’ll call that mass 𝑚 sub 𝑏.
As we work towards solving for 𝑚 sub 𝑏, let’s make an observation about this scenario. Our observation is that the seesaw is in equilibrium. That means that the sum of the forces acting on the seesaw is zero or that the net force on the seesaw is zero, and it also means that the net torque on the seesaw is zero.
Speaking of rotation, let’s define rotation in the clockwise direction as positive rotation. That means of course that rotation in the counterclockwise direction we’ll consider negative. To start solving for the mass of the board, let’s recall an equation for torque.
When the distance from the point of rotation to the applied force 𝑟 and the force 𝐹 are both in the same plane as they are in our situation, then torque is equal to their product times the sine of the angle between them. If we draw an expanded view of the seesaw, then we’re better able to see that there are three forces that act in the downward direction relative to the triangular fulcrum.
First, there’s the force of the boy on the left; then there’s the force of the board acting down. Since the board is three plus five or 8.0 meters long and is uniform, that force acts from the center of mass of the board one meter to the right of the fulcrum. 𝐹 sub 𝑔𝑏 is the gravitational force on the board. And finally, there is 𝐹 sub 𝑔𝑟, the gravitational downward acting force on the board on the right.
For each one of these forces, we can write them out in terms of the mass involved and 𝑔, the acceleration due to gravity. Each of these three forces exerts a torque about the fulcrum, tending to make the seesaw rotate either clockwise or counterclockwise.
In the equation for torque, 𝑟 is the distance between the fulcrum and where that force acts. By looking at the diagram, we can tell that 𝑟 sub 𝐿 is equal to 3.0 meters, 𝑟 sub 𝑏 is equal to 1.0 meters, and 𝑟 sub 𝑟 is equal to 5.0 meters. Since each one of these 𝑟 values is perpendicular to the force to which it is connected, we can disregard the sine of the angle between the two because the sine of that angle, 90 degrees, is one. This means we can write an equation for torque using these three terms.
Since we’ve defined clockwise motion as positive motion, the net torque acting around the fulcrum point is 𝑚𝑟𝑔 times 𝑟 sub 𝑟 plus 𝑚𝑏𝑔 times 𝑟 sub 𝑏 minus 𝑚𝐿𝑔 times 𝑟 sub 𝐿. And remember that since the seesaw is in equilibrium, this all is equal to zero.
We can rearrange this equation to solve for 𝑚 sub 𝑏, the mass of the board. First, let’s cancel out the common factor of the acceleration due to gravity, 𝑔. Next, let’s subtract 𝑚 sub 𝑟 𝑟 sub 𝑟 from both sides, canceling that term out on the left, and we’ll then add 𝑚 sub 𝐿 𝑟 sub 𝐿 to both sides, also canceling that term on the left.
Next, we divide both sides of the equation by 𝑟 sub 𝑏, canceling that term also from the left-hand side. This leaves us with an equation for the mass of the board, which says 𝑚 sub 𝑏 is equal to 𝑚 sub 𝐿 𝑟 sub 𝐿 minus 𝑚 sub 𝑟 𝑟 sub 𝑟, all divided by 𝑟 sub 𝑏.
Each of the five values on the right-hand side of this equation are given, so we can plug them in now. When we calculate this value, we find that 𝑚 sub 𝑏, the mass of the board, is 40 kilograms. That’s how much the board must weigh in order for the seesaw to be in equilibrium.