Your task in a science contest is to stack four identical uniform bricks, each of length L, so that the top brick is as far to the right as possible without the stack falling over. Is it possible, as FIGURE P12.60 shows, to stack the bricks such that no part of the top brick is over the table? Answer this question by determining the maximum possible value of d.

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Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a stacking tournament. Competitors are given three identical uniform rectangular wooden pieces each with a length capital L and a mass capital M competitors stack the wooden pieces on a horizontal platform aiming to have the top piece project as far as possible from the edge while maintaining the stacks stability. If the wooden pieces are stacked with their longest edge parallel to the platform, determine the maximum horizontal distance that the top most wooden piece can project from the edge of the platform. OK. So that is our end goal is to determine the maximum horizontal distance that the top most wooden piece can project from the edge of the platform. And this is shown in the diagram down below here on the right. It shows the three wooden blocks stacked on top of each other with their pieces as far as possible from the edge while maintaining stability. So it's basically they're trying to stack it just right. So they don't go toppling over. Ok. So we're given some multiple choice answers. Let's read them off to see what our final answer might be. A is 4/9 lb is 58 LC is 14, divided by 17 L and D is 11/12 L. Ok. So an L is the length just to remind ourselves. Ok. So first off, let us note that the first wooden piece from the top must have a center of mass that is exactly on the edge of the second, lower wooden piece to prevent it from falling over. Thus, the center of mass for the wooden piece must be L divided by two. OK. Therefore, the first wooden piece can extend past the second wooden piece by L divided by two. Now, we can consider what would happen if we place the 1st and 2nd wooden pieces on the edge of the third wooden piece, their combined center of mass must lie precisely over the tip of the third wooden piece. In order to prevent it from falling over, we can get, we can find the combined center of mass of the 1st and 2nd wooden pieces. And they can be determined by using the left tip of the second wooden piece and by treating it as the point of origin. So considering all of this, we could write that X is the distance, the maximum horizontal distance, the value that we're trying to find at the center of mass, which we're gonna denote it as subscript CM for center of mass for pieces, one and two is equal to M one multiplied by one plus M two multiplied by two divided by M one plus M two. OK. And let's note, let's make a little note up here that M one equals M two equals capital M. OK. So noting that let's rewrite this. So X CM 12 is equal to capital M multiplied by L divided by two plus capital, M multiplied by L divided by two plus L divided by two divided by capital M plus capital M. OK. So let's simplify this a little bit. So XCM 12. So the center of mass or the distance or the center of mass for pieces, one and two when we simplify will equal M multiplied by L divided by two plus capital M multiplied by L divided by two, multiplied by capital M. OK. So we can simplify once again. So the center of mass, the distance for the center mass for piece is one and two. When we simplify, once again, will equal 3/4 L thus the second wooden block can extend past the third block by L divided by four. So let's make a little note here. So to reiterate to say that again, so the second wooden block can extend past the third block by L divided by four. OK. So considering the entire stack, their combined center of mass must lie exactly over the tip of the platform for stability. So it doesn't fall over. We can determine the combined center of mass for the 1st, 2nd and 3rd wooden pieces. By considering from the left tip of the third wooden piece as the point of origin. So we need to find it by considering the left tip of the third wooden piece as the point of origin. So we can then go on to write that the distance from the center of mass for both the first, the second and the third wooden piece is equal to M one multiplied by one. So M one multiplied by one. So M one multiplied by one plus M two multiplied by two plus M three multiplied by three, right M three multiplied by three divided by M one and plus M one plus M three. OK. So let's start plugging in our nodes here and simplify this a little bit. So this distance for the center of mass for pieces, 12 and three is equal to capital M multiplied by L divided by two. As we discussed before plus capital, M multiplied by three L divided by four plus capital, M multiplied by five L five L divided by four plus capital M plus capital M. OK. So let's simplify it. So this distance for the center of mass for a piece is 12 and three is equal to 56 L. So five divided by six L. OK. So thus the third wooden piece can extend past the platform by L divided by six. Fantastic. So the maximum distance from the m platform edge to the right edge of the top piece can be written as OK. So it's called S Max. S. Max is equal to L divided by two plus L divided by four plus L divided by six. So once we get all the fractions and to have the same denominator and then we can add them together. And when, once we do that, our final answer will be 11/ L. So 11 divided by 12 L and that is our final answer. We did it. Ok. So that means looking at our multiple choice answers, the final answer and correct answer has to be the letter D divided by 12 L. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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Key Concepts

Center of Mass

Torque and Equilibrium

Static Friction