Torque equilibrium. Where would the masses need to be placed?

[johnball123]

A triple beam balance, like the scales we use every week in lab, works using rotational equilibrium. If a 253 g mass is placed on the pan, 1 cm from the pivot, and there are three masses, 1 g, 10 g, and 50 g masses on the other side can be slid back and forth between 0 cm and 10 cm from the pivot, where would the masses need to be placed? Full credit only if you provide values at whole-integers of centimeters for the three masses.

So far I've got: .253-.05r-.01(r+x)-.001(r+x+y)=0

but I can't figure out how to solve for all the variables. any help is appreciated!

 

[billy_joule]

Have you included the entire problem statement?
Your equation implies that maybe you haven't. You want your variables to be distance from the pivot, not from the previous mass, In your eqn 'x' is the distance from the 50g mass to the 10g mass, which isn't really relevant. So instead of your eqn:

0.253 -.005r - 0.01x - 0.001y = 0, where 0 ≤ r,x,y ≤ 10 and integers

This can be solved by iteration pretty quickly. There is one solution where they are three consecutive integers...

Or are you expected to be more mathematically rigorous?

 

[johnball123]

This is the whole problem statement. I apologize, I'm not sure what iteration is. I'm a liberal arts major out of my comfort zone, taking a college algebra-based physics course.

 

[gneill]

Since you'll be comparing moments about a pivot, so long as all the units are the same (grams for mass, cm for distances) you won't need to convert to kg and meters. Just write the sum of the torques for each side of the pivot as an equation. Place the torque for the 253 gram mass on one side of the equals and the sum of the other torques on the other side. You really don't even need to convert the masses to weights since that's just multiplying every mass on both sides by a constant (g), and thus they all cancel out.

The problem can then be solved quite easily by inspection.

 

[HalfLostAndSearching]

I can provide a response to your question about torque equilibrium. In order to achieve torque equilibrium, the total torque on both sides of the balance must be equal. This means that the clockwise torque must be equal to the counterclockwise torque.

In this situation, we have a 253 g mass placed 1 cm from the pivot on one side, and three masses (1 g, 10 g, and 50 g) on the other side that can be slid back and forth between 0 cm and 10 cm from the pivot. We can use the equation for torque, which is force multiplied by the distance from the pivot, to determine the placement of the masses.

First, we need to determine the total torque on each side. On the side with the 253 g mass, the torque is 253 g multiplied by 1 cm, which is 253 g*cm. On the other side, we have three masses, so we need to calculate the torque for each individual mass and then add them together. The torque for the 1 g mass is 1 g multiplied by the distance from the pivot, which is x cm. The torque for the 10 g mass is 10 g multiplied by the distance from the pivot, which is (x+y) cm. And the torque for the 50 g mass is 50 g multiplied by the distance from the pivot, which is (x+y+z) cm. Adding these together, we get a total torque of 1x+10(x+y)+50(x+y+z) g*cm.

Now, to achieve torque equilibrium, these two torques must be equal. So we can set up an equation:

253 g*cm = 1x+10(x+y)+50(x+y+z) g*cm

Simplifying this, we get:

253 g*cm = 61x+60y+50z g*cm

To solve for the variables, we need to have three equations. Since we have three variables (x, y, and z), we need three equations to solve for them. We can use the fact that the masses can be slid between 0 cm and 10 cm from the pivot to set up three equations.

At 0 cm from the pivot, the torque on the other side must be 0, so we can set up the equation:

0 = 1(0)+10(0+y)+50(0+y+z