## 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

1. Taufique

|dw:1332049690832:dw| the net force on the body =(9-8sin62)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. hosein

angle with normal or horizontal? if horizontal:$-8\cos62+90=ma$&$N+8\sin62=mg$

yeh if finding the xcoordinate use cos from what i know

4. hosein

ordinarily angle with horizontal|dw:1332050614025:dw|

i get 1.75m/s^2 but the answer is 2.9 m/s^2

6. Taufique

@hosein here Normal force will be equal to weight of the body.hence they will be cancel to each other.

i will try use pythagoras with vector notation

8. hosein

one the force has angle so it has component throgh normal

9. hosein

straight from the book...you don't consider gravity in this problem

11. hosein

why we use gravity !we have'nt got friction here so we don't use it

12. Taufique

@hosein,yes ,you can also do here but it is not necessary here

ah k i was just looking at your diagram...

14. hosein

2.9 m/s^2 and working is???

im still fiddling with angles to come up with an answer

ohh i see now

i think

19. hosein

i don't understand what are you saying

so 180deg mins 62degs =118degs so (9-8cos-118)/3=4.25 so (9-8sin118)/3=1.94 squareroot(4.25^2 + 1.94^2) = wrong answer arrrgh

21. Taufique

oh sorry the diagram is wrong , see true diagram here.|dw:1332051564007:dw| then net force =(9-8cos62)i^+(8sin62)j^ then follow same way for finding the acceleration. acceleration must be 2.9m/sec^2(about)