Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
PHYSICS HELP!!!!?
1.Three balls are attached to a light rigid rod, as shown in the figure. A fulcrum is located at xf = 19.0 cm measured from the left end of the rod. Ball A has mass mA = 40 g and is located at xA = 3 cm. Ball B has mass mB = 15 g located at xB = 22 cm. Ball C has mass mC = 24 g located at xC = 37 cm. Calculate the x-coordinate of the center of mass XCM. Assume that the mass of the rod is small enough to be ignored in your calculation. Your answers should be accurate to within 0.1 cm
the photo is just a rod with a fulcrum located near the center but more towards the left. Ball A is further left than the fulcrum. Ball B and Ball C are further right than the fulcrum with ball B closer to the fulcrum. All the length measurements are measured from the left of the rod
2. In Part 1 above, is the structure in balance? If not, how far must the fulcrum be moved so that it becomes in balance? (Answer "0.0" if it does not need to be moved. Also, signs matter: + if to the right, – if to the left.)
1 Answer
- Richie AlfredLv 75 years agoFavorite Answer
therefore CoM x-cordinate of object = [mᵢxᵢ + mᵢᵢxᵢᵢ + mᵢᵢᵢxᵢᵢᵢ] / [mᵢ + mᵢᵢ + mᵢᵢᵢ]
= [40(3) + (15)(22) + (24)(37)] / [40 + 15 + 24 ] = 1338/79 = 16.937 ~= 16.9 cm
anticlockwise moments = 40(g)13.9 = 556g
clockwise moments = 15(5.1)g + 24(20.1)g = 558.9g
if the fulcrum needs to be shifted right via a distance xₒ, to maintain equilibrium
=> 40(13.9 + xₒ)g = [15(5.1 - xₒ )g + 24(20.1 - xₒ)g]
=> xₒ = 558.9 - 556 = +2.9 cm
hope this helps
