Forces, vectors and acceleration

  • Thread starter netrunnr
  • Start date
  • #1
netrunnr
15
0

Homework Statement


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 8.2 N, acting 51° north of west. What is the magnitude of the body's acceleration?


Homework Equations



F=ma

The Attempt at a Solution



I have 51º NW at 8.2N and 0º E at 9N
51º is actually 141º from x-axis and I will do the calculations from there.

I am basically subtracting one vector from the other to see the resultant and I get the acceleration by using F = ma

math:
NWx = 8.2 cos 141º = -6.37i
NWy = 8.2 sin 141º = 5.16j

Ex = 9 cos 0º = 9i
Ey = 9 sin 0º = 0j

subtract to get
-6.37i+ 5.16j
+
9 i + 0j
______________________________
3.37i + 5.16j = R

now find R in N
3.372+5.162 = 37.98
[itex]\sqrt{37.8}[/itex] = 6.16N = F
6.16N = 3kg * a
6.16N / 3 kg = 2.05m/s2 = a

my answer is supposed to be 2.48m/s2
where did I lose it?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
121
subtract to get
-6.37i+ 5.16j
+
9 i + 0j
______________________________
3.37i + 5.16j = R

It looks like you're doing OK until you get to this bit. You say subtract, but you are adding the two vectors (which is the correct thing to do). But then the answer you get doesn't look right. You did -6.37+9 which is not equal to 3.37
 
  • #3
kushan
256
0
180-51= ?
 
  • #4
BruceW
Homework Helper
3,611
121
I think he's doing 90+51. This is another thing to watch out for. The question is a bit vague whether the angle is to be taken from the horizontal or from the vertical.
 
  • #5
kushan
256
0
oh , if you take the angle between the F vectors to be 129 , you get the answer correctly
 
  • #6
kushan
256
0
51 degree north of west means 51 degree taken from x axis in II quadrant??
 
  • #7
netrunnr
15
0
first thanks for pointing out the math error 9 - 6.37 is 2.63! my bad.

51 degree north of west means 51 degree taken from x axis in II quadrant??

yes, this is my problem. it was taken from the x axis in QII and I was taking it from the norther direction in Q II so I did 51º + 90º and it should have been 180º - 51º giving me an angle of 129º from the positive x axis.

thanks!
 

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