Am I missing important info on Torque? (TBR question)

[ilovemedi]

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A 1 kg meter stick is attached to the ceiling by a string at the 40 cm mark. If a 3 kg mass is attached to the meter stick at the end labeled 0 cm,what mass placed at the 70 cm mark would balance the meter stick?

Ans: 3.67 Kg

So I set the point where it rotates (at 40cm) as a point of reference. Thus:

Torque left = Torque right
3kg(40meters) = (mass trying to find)*g*30cm.

But the answer includes the torque by the meter stick so it adds "1kg*10*10cm" to the right side of the equation. 1) WHY include the torque of the ruler? do we always have to include the mass of a beam/rod/etc 2) why 10cm? and shouldn't the 10cm be added on BOTH sides of the equation then?

 

[jcb123]

So I solved this without the mass of the ruler and got 4 kg.

Where did the 6 come from in your calculation? that might be what threw you off.

Also, I include the torque of the ruler itself if the question mentions it. And it would only come in at the center of mass (so directly in the middle of the ruler at the 50 cm mark). You never want to include it on both sides. and this makes sense because the mass has a greater effect on the longer side from the string.

So if you solve with the mass of the ruler.

3(.4)10 - X (.3) 10 - 1(.1)10 = 0

X=3.67

 

[inasensegone]

A 1 kg meter stick is attached to the ceiling by a string at the 40 cm mark. If a 3 kg mass is attached to the meter stick at the end labeled 0 cm,what mass placed at the 70 cm mark would balance the meter stick?

Ans: 3.67 Kg

So I set the point where it rotates (at 40cm) as a point of reference. Thus:

Torque left = Torque right
3kg(40meters) = (mass trying to find)*6*30cm.

But the answer includes the torque by the meter stick so it adds "1kg*10*10cm" to the right side of the equation. 1) WHY include the torque of the ruler? do we always have to include the mass of a beam/rod/etc 2) why 10cm? and shouldn't the 10cm be added on BOTH sides of the equation then?

jcb pretty much answered your question, I just thought I would add the following, as it often helps to hear a couple different ways of explaining something.

1. In this case the problem told you that the mass of the meter stick is 1 kg. The only way that you can ignore this fact, is if the string was tied (fulcrum point) directly at the center of the meter stick (50 cm). In this case, the string is tied at the 40 cm mark, and so you must factor in the mass of the meter stick. The mass of the stick will factor in at the center of mass of the meter stick... i.e. 50 cm. (this is where you get the 1 kg * 10 N/kg * 10 cm).
2. The rest seems like you understand. Sum of Torque left = Sum of Torque right (although technically this is torque out of the page = torque into the page). Therefore you get 3 * 10 * 40 = 1 * 10 * 10 + MassUnk * 10 * 30. ---- "10" factors out so you could just say 120 = 10 + 30M ----> 110 = 30M ----> M ~ 3.67 kg

 

[ilovemedi]

jcb pretty much answered your question, I just thought I would add the following, as it often helps to hear a couple different ways of explaining something.

1. In this case the problem told you that the mass of the meter stick is 1 kg. The only way that you can ignore this fact, is if the string was tied (fulcrum point) directly at the center of the meter stick (50 cm). In this case, the string is tied at the 40 cm mark, and so you must factor in the mass of the meter stick. The mass of the stick will factor in at the center of mass of the meter stick... i.e. 50 cm. (this is where you get the 1 kg * 10 N/kg * 1 cm).
2. The rest seems like you understand. Sum of Torque left = Sum of Torque right (although technically this is torque out of the page = torque into the page). Therefore you get 3 * 10 * 40 = 1 * 10 * 10 + MassUnk * 10 * 30. ---- "10" factors out so you could just say 120 = 10 + 30M ----> 110 = 30M ----> M ~ 3.67 kg

Sorry I mean "g" instead of 6. And wow, I did not know we had to know a meter stick had a 100cm (I know that sounds super idiotic). I'm still confused at why it's not 1kg*g*50cm then, if 50cm is the center of mass of the ruler. Why 10cm? Because in another problem where we took the rod into consideration, we just left it at Length of Rod divided by 2 for the "r" (passage 2, question 8 of page 193).

 

[brood910]

Sorry I mean "g" instead of 6. And wow, I did not know we had to know a meter stick had a 100cm (I know that sounds super idiotic). I'm still confused at why it's not 1kg*g*50cm then, if 50cm is the center of mass of the ruler. Why 10cm? Because in another problem where we took the rod into consideration, we just left it at Length of Rod divided by 2 for the "r" (passage 2, question 8 of page 193).

That depends on where you set the pivot point.

 

[ilovemedi]

OH! Is it because you measure everything relative to the pivot point, which in this case is there the rope is attached to the ruler. So it's 50(center of mass of ruler) MINUS 40 (where the pivot point/rope is attached to) = 10 cm. Is that reasoning correct?