Video: Using Rotational Equilibrium Conditions about a Pivot to Determine the Mass of an Object

The uniform seesaw is balanced at its center of mass, as shown. The smaller boy on the right has a mass of 40 kg. What is the mass of his friend?

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Video Transcript

The uniform seesaw is balanced at its centre of mass as shown. The smaller boy on the right has a mass of 40 kilograms. What is the mass of his friend?

In this problem, we’re given the mass of the boy on the right side of the seesaw; that’s 40 kilograms.

And we want to solve for the mass of his friend; we’ll call that mass π‘š. To solve for that mass, we recognize that because the seesaw is balanced, that means that the net torque on the seesaw is zero.

In general, when an object does not exhibit motion, that implies that the torque on the object is zero, as it is here. So let’s calculate the torque that is exerted by each one of the two boys on the seesaw to solve for π‘š.

Let’s say that clockwise-direction motion is positive. And we’ll label the masses of the two boys; the mass of the boy on the left we’ve called π‘š and the mass of the boy on the right we’ll call π‘š sub π‘Ÿ.

Through the force of gravity on these two boys, each one exerts a torque around the fulcrum of the seesaw.

The boy on the right, π‘š sub π‘Ÿ, exerts a torque of his mass times gravity times the distance he is from the axis of rotation, which is given as four meters.

The boy on the left on the other hand exerts a torque in the opposite direction a distance of two point zero meters from the axis of rotation.

The overall torque on the seesaw we’ll call 𝜏. And we know that that is zero but it’s also equal to the sum of the torques provided by each boy, which we’ll call 𝜏 sub π‘Ÿ for the torque provided by the boy in the ride and 𝜏 sub 𝑙 for the torque created by the boy on the left.

So we now have an equation that reads zero is equal to 𝜏 sub π‘Ÿ minus 𝜏 sub 𝑙. Let’s recall the equation for torque in terms of force and distance. In general, torque is equal to π‘Ÿ, the vector from the axis of rotation, to where the force creating a torque is applied crossed with that force.

In our case, the angle between 𝐹 and π‘Ÿ, 𝐹 being the force created by the boys as mass and π‘Ÿ being a vector from the axis of rotation to those forces, that angle is always 90 degrees.

So we can write a simplified version of this vector equation for torque. In this scenario, torque 𝜏 equals π‘Ÿ times 𝐹.

We can now apply this fact to solve for 𝜏 sub π‘Ÿ and 𝜏 sub 𝑙. 𝜏 sub π‘Ÿ, the torque created by the boy on the right is equal to π‘Ÿ sub π‘Ÿ, the distance from the boy on the right to the axis of rotation, multiplied the force that boy exerts, namely the boy’s weight force.

We can therefore write 𝜏 sub π‘Ÿ as being equal to π‘Ÿ sub π‘Ÿ times π‘š sub π‘Ÿ times 𝑔. We can do something similar for 𝜏 sub 𝑙 the torque created by the boy on the left 𝜏 sub 𝑙 is equal to π‘Ÿ sub 𝑙, the distance from the rotation point to the boy, multiplied by 𝐹 sub 𝑙, which in this case is the weight force of the boy.

We can therefore write this expression as π‘Ÿ sub 𝑙 times π‘š times 𝑔. Let’s take these expansions of the terms 𝜏 sub π‘Ÿ and 𝜏 sub 𝑙 and insert them into our equation for the torque balance.

So we now have an expression that reads zero is equal to π‘Ÿ sub π‘Ÿ π‘š sub π‘Ÿ 𝑔 minus π‘Ÿ sub 𝑙 π‘šπ‘”. If we divide both sides of the equation by 𝑔, the acceleration due to gravity, that term cancels out of the equation leaving us with the expression zero is equal to π‘Ÿ sub π‘Ÿ π‘š sub π‘Ÿ minus π‘Ÿ sub 𝑙 times π‘š.

Let’s add π‘Ÿ sub 𝑙 times π‘š to both sides of the equation, which results in that term cancelling out of the right side of our equation.

If we now divide both sides of the equation by π‘Ÿ sub 𝑙, which cancels that term on the left-hand side of our equation, we’re left with an expression for π‘š, the mass of the boy on the left.

That boy’s mass is equal to π‘Ÿ sub π‘Ÿ, the distance from the boy on the right to the axis of rotation, multiplied by his mass π‘š sub π‘Ÿ divided by π‘Ÿ sub 𝑙, the distance from the boy on the left to the axis of rotation.

As we look on the diagram, we see that π‘Ÿ sub π‘Ÿ is equal to four point zero meters and that π‘Ÿ sub 𝑙 is equal to two point zero meters. And furthermore, we’re given that π‘š sub π‘Ÿ, the mass of the boy on the right, is 40 kilograms.

We then enter these three values into our equation to solve for π‘š, which equals 4.0 meters times 40 kilograms divided by 2.0 meters.

And when we enter those numbers into our calculator, we find a value of 80 kilograms. That’s the mass that the boy on the left would have in order to balance out the seesaw.

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