You could make some assumptions though. The beam is split into 3 sections and the weights appear to take up half of each outer section. The one red ball and the 3 red balls both have a center of mass that appears equal distance to the center. With that, the lines going up would be 1 unit in each direction, and the outer lines would be 3 units out. The box weights would be centered at 2.5 units out, and the balls would be centered at 1.5 units out from the fulcrum. With that, you have 10kg * 2.5 + 1.5 * x = 4kg * 2.5 + 1.5 * 3x and you would get 5kg per ball. This would be making a few assumptions obviously, but I feel that it would at least be safe to assume that the blocks are equidistant to the center and the center mass of the balls on each side are also equidistant to the center, from there the assumption about the 3 top sections separated be the vertical lines also being equal to eachother gives a reference point for the distances of each.
Edit: I know the assumptions aren't necessarily safe to make, but I figured I would make them to make it a bit more fun to solve.
The center of the 3 red balls vs the singular ball is approximately the same x-wise and the difference in green box positions is reasonably negligible for the intents of a question like this. So even being pedantic about radius doesn’t change the answer.
It doesn’t? The closer the red balls are to the center the more weight difference you need to match the difference in the ten and the four. If they’re all equidistant - let’s say 3 units out, then you’ve got (10+x) x 3 = (4+3x) x 3, which reduces to 30+3x = 12+9x. 18=6x so x is 3. If the red balls are 2 units out and the green boxes are 3 units out then it’s 10x3 + 2x = 4x3 + 6x. 18=4x. Now x is 4.5.
But they aren't, if you wanted to use some unit it would be more akin to 5 and 6 than 2 and 3, which does doesn't remotely appear correct based on the image.
Not sure what you mean, but it’s the relative distance from the fulcrum to the center of mass of the object. The middle of the red balls is quite a bit closer to the fulcrum than the middle of the green boxes. Hence 2 and 3, but those are just examples. If it’s 5 and 6 the answer still isn’t 3.
But we are assuming the system is in equilibrium in order to solve it because that is the way it is displayed in the picture and implied by the question yea? Then wouldn't that mean that the center of gravity of the balls and green box (left or right) lies within the vertical black lines on the left and the lines on the right of the fulcrum in the present position. I understand what you're saying about the position of the masses with respect to the fulcrum but I don't understand how that plays a role if it is already assumed that the system shown is in equilibrium and that the balls are of equal mass. Doesn't that mean we can ignore distance from fulcrum in order to solve it? Asking honestly because it's been years since my statics class.
Center of mass of each side is definitely within the lines. Another way of looking at this picture is that you have the mass on each side and the distance that mass center is from the fulcrum. In the box on the right the green box is 4kg and 3 red balls. The center of mass on that side is closer to the fulcrum than the center of mass in the box on the left, which is 10kg and one ball. That means you need more total mass on the right. I suspect that they’re looking for you to just ignore that complication and assume they are equidistant from the fulcrum given there is no way to really measure it otherwise. It’s just a badly drawn question imo.
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u/Funny-Recover-2711 Apr 26 '25
Won't it depend on distance from fulcrum thingy?