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anonymous
 4 years ago
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62° north of west. What is the magnitude of the body's acceleration?
anonymous
 4 years ago
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62° north of west. What is the magnitude of the body's acceleration?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1332049690832:dw the net force on the body =(98sin62)i^+(8cos62)j^ Apply vector method to find the magnitude of the net force. after it find the acceleration: acceleration=(magnitude of the net force)/mass of the body.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0angle with normal or horizontal? if horizontal:\[8\cos62+90=ma \]&\[N+8\sin62=mg\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeh if finding the xcoordinate use cos from what i know

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ordinarily angle with horizontaldw:1332050614025:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get 1.75m/s^2 but the answer is 2.9 m/s^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@hosein here Normal force will be equal to weight of the body.hence they will be cancel to each other.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i will try use pythagoras with vector notation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0one the force has angle so it has component throgh normal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my ans is same to you @jadefalcon are you sure about answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0straight from the book...you don't consider gravity in this problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why we use gravity !we have'nt got friction here so we don't use it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@hosein,yes ,you can also do here but it is not necessary here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah k i was just looking at your diagram...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the answer is..........?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02.9 m/s^2 and working is???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im still fiddling with angles to come up with an answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't understand what are you saying

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 180deg mins 62degs =118degs so (98cos118)/3=4.25 so (98sin118)/3=1.94 squareroot(4.25^2 + 1.94^2) = wrong answer arrrgh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh sorry the diagram is wrong , see true diagram here.dw:1332051564007:dw then net force =(98cos62)i^+(8sin62)j^ then follow same way for finding the acceleration. acceleration must be 2.9m/sec^2(about)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahhh yes he is right...but you forgot to divide by 3 at the end
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