Ek physics q

[inaccensa]

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Can anyone please explain how the mass is 2kg. I cant seem to understand Ek's solution..many thanks

 

[MD Odyssey]

Can anyone please explain how the mass is 2kg. I cant seem to understand Ek's solution..many thanks

Consider the torques about the pivot point. There are two torques:

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Solve for the mass of the bar and you can see that the mass is equal to 2 kg. Pretty straightforward.

 

[docntrainin]

The 0.3 lever arm is the distance from the T of the rope to center point of the board, correct? Can you please explain why. Thanks!

 

[MD Odyssey]

The 0.3 lever arm is the distance from the T of the rope to center point of the board, correct? Can you please explain why. Thanks!

For uniform density problems, such as this one, the mass of the rod acts at its center of mass. The center of mass, for a uniform rod is in the center, which is 0.3 meters from the pivot point, which is the point I chose to take torques about. In general, one can take torques about any point, but I tend to use the pivot point for problems like this.

Make sense?

 

[inaccensa]

I was trying to equate the mass with the the tension at no avail. If I had selected the pivot point where the 3kg mass is hung, the torque will be zero right? but lets say, if i want to use that point as a pivot, how will i solve this q? or i cannot?

 

[MD Odyssey]

I was trying to equate the mass with the the tension at no avail. If I had selected the pivot point where the 3kg mass is hung, the torque will be zero right? but lets say, if i want to use that point as a pivot, how will i solve this q? or i cannot?

It really depends upon the point you wish to take torques about. Clearly, the tension in the rope is just going to be the weight of the bar. But, since you don't know the weight of the bar, that's not all that helpful.

The problem is trying to see whether or not you will be distracted by the additional information of the tension in the rope. In general, the best way to solve these types of problems is to take your torques about the pivot point, which would have been the point where the string is tied. This problem is pretty easy - imagine a 1 meter long wooden bar. Drill a hole at one end and hang a weight there. Measure over 20 cm and tie your string there. Now, if you're told that the bar is in equilibrium, you know that the sum of the torques is zero. From there, you can solve for the mass of the bar directly.

 

[inaccensa]

It really depends upon the point you wish to take torques about. Clearly, the tension in the rope is just going to be the weight of the bar. But, since you don't know the weight of the bar, that's not all that helpful.

The problem is trying to see whether or not you will be distracted by the additional information of the tension in the rope. In general, the best way to solve these types of problems is to take your torques about the pivot point, which would have been the point where the string is tied. This problem is pretty easy - imagine a 1 meter long wooden bar. Drill a hole at one end and hang a weight there. Measure over 20 cm and tie your string there. Now, if you're told that the bar is in equilibrium, you know that the sum of the torques is zero. From there, you can solve for the mass of the bar directly.

Thanks a lot. it really helped