Balancing Three Masses First mass on left side m1

Balancing Three Masses First mass on left side m1 | Lever arm l1 | Second mass on left side m2 | Lever arm l2 | Mass on right side m3 | Lever arm l3 20g | 10 cm | 30g | 15 cm | 40g | 17 cm 20g | 20 cm | 30g | 16 cm | 40g | 18 cm 17. The net torque on left side of the ruler can be found by adding the individual torque (t_left = t_left1 + t_left2) for each mass on the left side. Calculate the torque on the left and right side of the ruler for each trial. 18. Calculate the percent difference between the torques on each side of the ruler for each trial. Is the percent difference small enough to conclude that the torque on each side of the ruler is equal?

Transcript

00:01 In this question, we are given some masses that were hung on a ruler, balanced at its center, and we're asked to find, and we're also given the distances from that balancing point, and we're asked to find the net torque on the left and right side of the balancing point, and then to calculate the percent difference to see if we can call these values experimentally the same.

00:28 And our instructions say to take, to get the torque on the left to calculate the individual torques and add them.

00:44 So we can have the torque on the left will be the torque from the first mass plus the torque from the second mass, which is gonna be m1g times l1 plus m2g times l2, because we need the force that's applied, which means we're gonna need our masses in kilograms.

01:09 And when you're calculating torque, there's also usually a sine theta for the angle between the force and the distance from the axis, but it's gonna be 90 degrees in this case, so we'll just let that be, because the sine of 90 is one.

01:25 So expressing our masses in kilograms, 0 .020 times 9 .8.

01:34 We also need our distances in meters, so multiplied by 0 .1 plus 0 .030 times 9 .8 times 0 .15.

01:48 And i get a torque for the left side of 0 .0637 newton meters.

01:57 And the torque for the right is a little bit simpler because we only have one object.

02:04 So we'll take m3gl3, and m3 was 0 .040 kilograms times 9 .8 times 0 .17 meters for its distance.

02:18 And it gives me 0 .0664 newton times meters.

02:28 And now for our second trial, we're going to do the same thing.

02:38 In fact, i think i probably have enough room to put that up a little bit.

02:48 So the torque on the left will once again be m1gl1 plus m2gl2.

02:57 So in this case, we have 0 .020 times 9 .8.

03:05 And now it has a distance of 0 .20 meters plus 0 .30.

03:13 Fix that, that zero looks like a six.

03:17 0 .030 times 9 .8 times, now it has a distance of 10 centimeters, 0 .10 meters.

03:27 And i get a torque on the left of 0 .0686 newton meters.

03:34 Newton times meters.

03:36 And now for the torque on the right, once again, we'll take m3gl3.

03:46 So 0 .040 kilograms times 9 .8 times 0 .16 meters gives me 0 .06272 newton times meters...

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