physics_0
2017/07/04 04:04
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標題:
physics
發問:
Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25degrees west of north, and the resultant of thesetwo pulls is 350.0N directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the smaller... 顯示更多 Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25degrees west of north, and the resultant of thesetwo pulls is 350.0N directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull. http://www.sendspace.com/file/qbxgit I draw the direction and vectors, am I right? Please list the steps Ans:149 N 32.2 N of E
最佳解答:
Let F be the magnitude of the smaller force, which makes an angle of a with the resultant force of 350 N. Hence, resolve the forces into components in the north and east-west directions. In the east-west direction, the components are in equilibrium: 2F.sin(25) = F.sin(a) i.e. a = arc-sin[2 x sin(25)] degrees = 57.7 degrees In the north direction: 2F.cos(25) + F.cos(57.7) = 350 i.e. F = 350/(2 x cos(25) + cos(57.7)) N = 149 N Therefore the two forces are 149 N and 2x149 N = 298 N The 149 N force makes an angle of (90 - 57.7) degrees = 32.3 degrees north of east.
physics
發問:
Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25degrees west of north, and the resultant of thesetwo pulls is 350.0N directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the smaller... 顯示更多 Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25degrees west of north, and the resultant of thesetwo pulls is 350.0N directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull. http://www.sendspace.com/file/qbxgit I draw the direction and vectors, am I right? Please list the steps Ans:149 N 32.2 N of E
最佳解答:
Let F be the magnitude of the smaller force, which makes an angle of a with the resultant force of 350 N. Hence, resolve the forces into components in the north and east-west directions. In the east-west direction, the components are in equilibrium: 2F.sin(25) = F.sin(a) i.e. a = arc-sin[2 x sin(25)] degrees = 57.7 degrees In the north direction: 2F.cos(25) + F.cos(57.7) = 350 i.e. F = 350/(2 x cos(25) + cos(57.7)) N = 149 N Therefore the two forces are 149 N and 2x149 N = 298 N The 149 N force makes an angle of (90 - 57.7) degrees = 32.3 degrees north of east.
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