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jadefalcon
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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?
 2 years ago
 2 years ago
jadefalcon Group Title
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?
 2 years ago
 2 years ago

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Taufique Group TitleBest ResponseYou've already chosen the best response.0
dw: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.
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
angle with normal or horizontal? if horizontal:\[8\cos62+90=ma \]&\[N+8\sin62=mg\]
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
yeh if finding the xcoordinate use cos from what i know
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
ordinarily angle with horizontaldw:1332050614025:dw
 2 years ago

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

Taufique Group TitleBest 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.
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
i will try use pythagoras with vector notation
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
one the force has angle so it has component throgh normal
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
my ans is same to you @jadefalcon are you sure about answer?
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
straight from the book...you don't consider gravity in this problem
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
why we use gravity !we have'nt got friction here so we don't use it
 2 years ago

Taufique Group TitleBest ResponseYou've already chosen the best response.0
@hosein,yes ,you can also do here but it is not necessary here
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
ah k i was just looking at your diagram...
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
so the answer is..........?
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
2.9 m/s^2 and working is???
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
im still fiddling with angles to come up with an answer
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
ohh i see now
 2 years ago

hosein Group TitleBest ResponseYou've already chosen the best response.2
i don't understand what are you saying
 2 years ago

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

Taufique Group TitleBest ResponseYou've already chosen the best response.0
oh 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)
 2 years ago

jadefalcon Group TitleBest ResponseYou've already chosen the best response.0
ahhh yes he is right...but you forgot to divide by 3 at the end
 2 years ago
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