A 0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass gl...

Question

A 0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.110-kg mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

Accepted Answer

Hey everyone in this problem. A student glues a 45 g mass at the left extremity of a one m bar and a 15 g mass. 10 centimeters from the right, extremely OK. The small masses are considered point masses. The bar mass is 500 g and it's uniformly distributed along its length. The student is asked to balance the mass bar combination horizontally on a pivot situated precisely below the combination center of gravity. Okay. And were asked at what distance from the right extremity must the pivot be placed? Okay, Alright, so let's draw this out. We have our bar. Hey, sitting horizontally, we have the mass at the left end, we have a mass that's 10 centimeters from the right end. We know that the length of our bar is one m. Alright, so this is what our picture looks like and what we could be thinking here is okay, this is an equilibrium situation. We have no net external force is we have no net external torques. So let's use our equilibrium conditions where the sum of the forces is zero. The sum of the torques is equal to zero to solve this problem. Okay. And we can draw out the forces. Okay, we would have the weight l from the left mass. We would have the weight are from the right mass. Okay, this bar is uniformly distributed, so its center of mass will be right in the middle. Okay? And we'll get the weight from the bar and then we're gonna place a pivot and so we're going to have that force acting upwards from that pivot. Okay. And we could use those equal room conditions to find where this FP should be located but we can actually do something simpler. Okay, solving it that way. We'll get you the right answer. But we're gonna do something simple. Okay, these are point masses. Okay? We want to place the pivot at the center of gravity so let's just go ahead and calculate the center of gravity through the center of gravity equation. Okay now when we're talking about the center of gravity or the center of mass, what masses do we need to consider? Okay, well we have the mass on the left side and the mass on the right side. And then we also have the mask from the bar. Mhm. So our equation for the center of mass X cm is going to have three components. So we have M one X one plus M two X two plus M three X three. All divided by the sum of the masses M one M two and M three. Alright, so let's start with This equation and we're just going to rewrite it in terms of the variables that we have. Okay, so let's consider M1 to be the mass of the left mass. Okay. And then its position or its distance L M two is going to be the mass of the bar. Okay? And then the variable was subscript three. We're gonna call M sub R. Okay. And this is gonna be relating to that right mass divided by the sum of the three M L M B and M are Okay. All right now we want to know the distance from the right extremity that we have to put this pivot. So let's choose the right as our reference point. Okay? So when we're talking about distances, it's always going to be from that right point. So the mass of this left Mass is 45 g. Okay? And it's at the end, the left end. So this is going to be 45 g times now the distance and we're taking the distance from the right end. Okay, This is going to be our reference point. So the distance from that red point to our left is the distance of the bar or the length of bar, which is one m. Okay, We have plus the second mass or the mass of the bar, which we know is 500 g in the distance of the center of the mass of that bar. Okay, because it's uniformly distributed, the center of mass is right in the middle. And so the distance to its center of mass is 0.5 m. Okay? And then we have the right mass. We're told that this is 15 g. Okay. And it's at a distance of 10 cm. Okay, so 10 cm we want to convert to meters. So we multiply by one m per 100 centimeters. We get that this is equal to 0.1 m. So 15 g times 0.1 m. And then we're going to divide by the sum of the masses. 45 g plus 500 g plus g. Okay? And we get a center of mass X cm 0.5 to nine m. And we look at our answer choices. These are in centimeters. Okay? So let's convert two centimeters to convert two centimeters. We take 0.5 to nine m. We multiply by 100 centimeters per meter. Okay? The unit of meter will cancel and we get 52. centimeters. That's going to be the center of mass Of this system, which is where we want to place that pivot. Okay. And so the pivot should be placed at D 52.9 cm from the right end. Thanks everyone for watching. I hope this video helped see you in the next one.

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