A plank of wood of length 400m with a fulcrum pivot at its...

A plank of wood of length 4.00 m with a fulcrum (pivot) at its center forms a seesaw. Suppose a child of mass 30.0 kg sits 1.50 m to the left of center, and a second child sits at the end on the far right side. The system is in equilibrium. Find the mass of the second child. (options: 30.0 kg, 22.5 kg, 7.5 kg, 40.0 kg) For the same setup with the two children sitting on the seesaw used in question 4, except now the first child's mass is 33.333 kg and the second child's mass is known to be 25 kg, find the normal force acting at the pivot point (fulcrum). The mass of the seesaw is 12.5 kg. (Hint: Remember if you can show the system is in equilibrium, then you can set the sums of forces/torques to zero and solve for whichever quantity you want.) Options: 804 N, 515 N, 695 N, 486 N

Transcript

0:00 Hello.

00:01 For this problem, you're considering a seesaw with two children on it.

00:08 I've gone ahead and drawn a diagram of the system and labeled the information that was given in the problem.

00:15 We're told for this first problem that the system is an equilibrium.

00:24 So that means that the sum of the torques is equal to zero and the sum.

00:30 Sum of the forces is equal to zero.

00:34 And we want to find the mass of the second child.

00:39 I have decided that the easiest way to go about this will be to use the equilibrium condition that the sum of the torques is equal to zero because we can see that if we take the fulcrum as our pivot point, the only torques we have to account for is the torques.

01:00 From the first child and the torque from the second child.

01:08 So for our first child, the torque is going to be a positive torque, and the magnitude will be equal to m1r1.

01:23 For the second child, the torque will be a negative torque, and the magnitude will be equal to m2r2.

01:36 And then, of course, that must be equal to zero.

01:39 So if we solve for m2, we see that m2 must be equal to m1r1 divided by r2.

01:49 And now plugging in the information from the problem, so 30 kilograms for the mass of child 1, 1 .5 meters for, the distance from the fulcrum for child one and two meters for the distance for child two, you end up with a mass of about 22 .5 kilograms for child two.

02:16 Now for the second part, we're told to consider some different things.

02:22 So i'm going to switch to red.

02:24 We're told that the setup is the same, but now our first child is no longer three.

02:30 30 kilograms, they are 33 .333 kilograms in mass.

02:40 And the second child's mass is known.

02:43 The second child has a mass of 25 kilograms in this case.

02:49 But everything else is the same.

02:51 So we are going to assume that r1 and r2 are the same.

02:55 We are additionally given the mass of the seesaw itself, which i've labeled in my diagram as m, and we're told that it should be 12 .5 kilograms.

03:09 We're asked then to find the normal force acting at the pivot point or the fulcrum.

03:16 And so now we want to think about this second equilibrium condition, because we know we'll have mass for child one, the mass for child two, and the mass of the seesaw, all directed, downwards and then the normal force directed upwards.

03:38 So if we use the forces in the y direction, that should work...

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