Question
In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.350 m (1.15 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 12.00 millibar. The atmospheric pressure outside is 960 millibar. Calculate the force required to pull the two hemispheres apart.
Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1330N (i.e., about 299 lbs), what is the minimum number of horses required?
Chrome File Edit View History Bookmarks People Window Help . D chem252 2nd review hanc xe"Sign in to your account LLON CAPA Magdeburg he ×'eAPiano Emits Sound wa You x https://sl lite.msu.edu/res/msu/kashy/Champ2002/champprobs/Magdeburg problem?symb-uploaded%2fmsu%2f6i31 52883c582567dmsull%2fdefault 1. » Homework Set #7 (due 3/22 at 11:59PM) » Magdeburg hemispheres Timer Notes à Evaluate Feedback- Course Contents » Print Info In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.350 m (1.15 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 12.00 millibar. The atmospheric pressure outside is 960 millibar. Calculate the force requíred to pull the two hemispheres apart. [Note: 1 millibar-100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ] Submit Answer Tries 0/20 Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1330N (i.e., about 299 lbs), what is the minimum number of horses required? Submit Answer Tries 0/20 readed View Chronological View Other Views_.. My general preferences on what is marked as NEW Mark NEW posts no longer new NEW 2nd question help Brianna Maria Canedo Reply (Wed Mar 9 11:52:26 pm 2016 (EST)) what am i doing wrong, so i take the first answer which i got 404709 and i divide that by 1480 then i multiply it by 2 but it keeps saying i have the wrong answer, not sure what else to try NEW Re: 2nd question help Richard Hallstein Reply (Wed Mar 16 07:14:37 pm 2016 (EDT)) While fractional horses exist, they tend to be dead and cannot pull. You should round up to the nearest even number of horses My settings for this discussion: 1. Display - All posts 2. Not new -Once marked not NEW Change Threaded View Chronological View Other Views.... Export Ehroaded View Chronological View other Views My general on what is marked as NEW Mark NEW posts no longer new nce
Explanation / Answer
According to the question we need to find the force required to pull the two hemispheres apart and minimum number of horses required.
Area = pie * (r) ^2 = 3.14 * ( 0.350 m) ^2 = 0.38465 m^2
Pressure(diff) = 960 millibar - 12.00 millibar = 948 millibar
Convert millibar to N/m^2 = 948 * 100 = 94,800N/m^2
Solve for force = 94,800N/m^2 * ( 0.38465 m^2 ) = 36464.82 Newton ============Part a)
Part b) No of horses = ( 36464.82 Newton ) / ( 1330N) = 27(each side) = 27 * 2 = 54 horses ====Part b)