A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw of FIGURE EX12.32. How far to the left of the pivot must a 4.0 kg cat stand to keep the seesaw balanced?

Question

Accepted Answer

Hey, everyone in this problem, a 5 m long plank rests on a F and a pivot or mid point on one end of the plank, we have a 4 kg backpack, right? That's on the right end. If we're looking at our diagram on the opposite end, there's a 1.5 kg lunch box. OK. In order to maintain equilibrium on the planet, we're asked to determine the distance at which a 2.5 kg rack should be positioned to the left of the planks mid plan. OK? So looking at our diagram, we're looking for this distance deep and that's the distance from the midpoint to the rat on the left. We have four answer choices here all in meters. Option A 2.5 option B 1.5 option C 0.5 and option D one. Now, in order for this to be maintained in equilibrium, OK. We need the center of mass to be at this fulcrum or pivot point which is at the mid. OK. So we know what the center of mass position should be. So let's go ahead and use our center of mass equation and we're gonna take the center of mass from the left-hand end. And that's gonna be our reference point. And I've put that in red and I'm gonna draw a star there. Ok. So what we called the position or center of mass, which we're gonna call XCM, it's gonna be equal to, well, the position of the lunch bag and relative to a reference position multiplied by the mass of the lunch bag. What's the position of the rat relative to that position? Multiplied by the mass of the rat plus the position of the backpack. Again, relative to our reference point multiplied by the mass of the backpack all divided by the sum of the masses ML plus Mr plus M. OK. So we're using subscript L to indicate lunchbox, subscript R for rat and subscript B for Yeah, our again, we want it to be at this green dot right in the middle of our plank, we have a 5 m long plank. So it's gonna be 2.5 m from the left end. The position of our center of mass is gonna be 2.5 m. Again, taking everything relative to our reference point on the far left. So 2.5 m is going to be equal to. Now the distance between our lunchbox a reference point. Well, they're both on the left end but the distance between the two is just zero, that relative position is zero. And so if XL is gonna be 0 m multiplied by the mass of 1.5 kg. And moving to the next part, we have the position of the ra Xr. Now, we wanna find Xrxr is gonna be related to that distance. D So Xr is what we're looking for here, multiplied by the mass of the rat 2.5 kg. What's the position of the backpack or the distance of the back? Now again, we're taking the left end to be a reference point. So the distance from that point to our backpack is the full length of that plank which is m. And that's gonna be multiplied by the mass of the backpack. 4 kg divided by the sum of the masses of those three objects. So we have 1.5 kg. I lost 2.5 kg last 4 kg. All right. So now we have an equation with only one unknown xr we can go ahead and simplify and solve. If we simplify the denominator on the right hand side, 1.5 plus 2.5 plus four, that's gonna give us 8 kg. We can multiply both sides by 8 kg. We have 2.5 m multiplied by kg is gonna be equal to. Well, the first turn on the right hand side is gonna go to zero. So we don't have to worry about that. On the right, we have xr multiplied by 2.5 kg. So what 20 kg meters? Ok. 5 m multiplied by four kira. All right. So we want to stop for Xr. So let's move this 20 kg meters to the other side, we have Xr multiplied by 2.5 kg is gonna be equal to, well, 2.5 multiplied by eight is also 20. So we get 20 kg meters minus 20 kg meters. That's just gonna be zero. So Xr multiplied by 2.5 kg just gonna be equal to zero, which tells us that Xr must be equal to 0 m. All right, so Xr is equal to 0 m. So we have to stop and be a little bit care that is not the distance we were looking. Ok. Let's go back up to our diagram and I'll show you what I mean. So let's remember that the distance we just found Xr is the distance from our reference point which is the end, the left end. OK? So we want the distance from that left end to a rat to be 0 m. That means we want a rat to be on the far left end. And so the distance deep from the center to the rat is gonna be 2.5 m. Ok? So the correct answer here is gonna be option A now this makes sense. Ok. The mass of that backpack is 4 kg, the mass of the lunch box 1.5 kg and the mass of the rat 2.5 kg combined is gonna be 4 kg. So if we put the rat right where that lunch box is, then the backpack on one side in the lunchbox and rat are gonna balance each other out right in the middle on that. Thanks everyone for watching. I hope this video helped see you in the next one.

A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw of FIGURE EX12.32. How far to the left of the pivot must a 4.0 kg cat stand to keep the seesaw balanced?

Verified Solution

Key Concepts

Torque

Center of Mass

Equilibrium