torque

[pizza1994]

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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?

source: TBR

how do you do the math for this? 🙁

 

[t5Nitro]

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?

source: TBR

how do you do the math for this? 🙁

So you need to consider the center of mass of the meter stick which is at the 50 cm mark at the center. Draw this diagram out, label one end 0 cm and the other end of the meter stick 100 cm with a string holding it up at 40 cm mark. At 50 cm is the center of mass where you can call it a 1kg weight.

You want the torque on the left side to equal the torque on the right side of that string holding it up.

On the left side, you have a 3 kg weight at the 0 cm mark (which is 40 cm away from the string). On the right side, you have the 1kg mass of the meter stick (located 10 cm from the string since it is at 40 cm). Further to the right is an unknown mass at 70 cm (located 30 cm from the string).

Torque left (3 kg mass) = torque right (meter stick mass + unknown mass)

Torque = Fd, and force is mass x gravity

Left side: (3 kg)(10 m/s/s)(0.40 m)

Right side: (1 kg)(10 m/s/s)(0.10 m) + (unknown mass)(10 m/s/s)(0.30 m)

Edit: I converted the distances into meters, but you wouldn't have to since they cancel out.

Set them equal: 12 = 1 + (unknown)(3) ---> 11 = 3x ---> x = 3 and 2/3 or 3.67 kg.

 

[pizza1994]

So you need to consider the center of mass of the meter stick which is at the 50 cm mark at the center. Draw this diagram out, label one end 0 cm and the other end of the meter stick 100 cm with a string holding it up at 40 cm mark. At 50 cm is the center of mass where you can call it a 1kg weight.

You want the torque on the left side to equal the torque on the right side of that string holding it up.

On the left side, you have a 3 kg weight at the 0 cm mark (which is 40 cm away from the string). On the right side, you have the 1kg mass of the meter stick (located 10 cm from the string since it is at 40 cm). Further to the right is an unknown mass at 70 cm (located 30 cm from the string).

Torque left (3 kg mass) = torque right (meter stick mass + unknown mass)

Torque = Fd, and force is mass x gravity

Left side: (3 kg)(10 m/s/s)(0.40 m)

Right side: (1 kg)(10 m/s/s)(0.10 m) + (unknown mass)(10 m/s/s)(0.30 m)

Edit: I converted the distances into meters, but you wouldn't have to since they cancel out.

Set them equal: 12 = 1 + (unknown)(3) ---> 11 = 3x ---> x = 3 and 2/3 or 3.67 kg.

why isnt center of mass of 1 kg at the 40 cm mark...i would put it there becasue thats where the string holds the rod!!!!!

 

[t5Nitro]

Think about it. If you put a string at 40 cm on a 100 cm stick, then it isn't going to sit level. That is not the center of mass.

Where the string is attached you have a 'left' and 'right' relative to where the string is placed. The torques on the left have to equal the torques on the right to keep it from rotating, and the meter stick weight is 'to the right' (if you drew the diagram with 0 cm on the left side of your paper) in this case.

 

[pizza1994]

Think about it. If you put a string at 40 cm on a 100 cm stick, then it isn't going to sit level. That is not the center of mass.

Where the string is attached you have a 'left' and 'right' relative to where the string is placed. The torques on the left have to equal the torques on the right to keep it from rotating, and the meter stick weight is 'to the right' (if you drew the diagram with 0 cm on the left side of your paper) in this case.

alright what you say makes sense! I think I get it and hopefully the next question I will get right haha thanks 🙂