Forces and acceleration on frictionless table?

In summary, the two forces F1 and F2 acting on a 27.0-kg object on a frictionless tabletop, with F1=10.2 N and F2=16.0 N, can be broken down into components and added together using trigonometric relations. The net force on the object and its acceleration can be calculated for cases (a) and (b), where (a) is at 90 degrees in the third quadrant and (b) is at 120 degrees from the positive y-axis and moving clockwise. To find the angle at which the net force acts, start at the positive x-axis and move counterclockwise until the angle is reached. The magnitude of the acceleration can be calculated using
  • #1
syavian1019
6
0
The two forces F1 and F2 act on a 27.0-kg object on a frictionless tabletop. If F1=10.2 N and F2=16.0 N find the net force on the object and its acceleration for cases (a) and (b).

(a) is 90 degrees, in the 3rd quadrant
(b) is 120 degrees, from the positive y-axis and moving clockwise

To calculate the angles at which the resultant forces act in this problem, please start at the positive x-axis and travel counter-clockwise.

1. The net force in figure (a) (above) is?

2. The angle from the positive x-axis at which the net force in figure (a) acts is?

3. The magnitude of the acceleration of the object in figure (a) is?

-----------------​

For 1, I don't know which formula I'm supposed to use. I can't find it in my textbook or anywhere else... but I think for 3 it's: a = (net force)/(mass in kg)

For 2, is this how you do it?: from the positive x-axis, move counterclockwise until you hit the angle.

Thanks for all help!

(Sorry about the broken links)
 
Last edited:
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  • #2
The link to the picture doesn't work (you need an account on that website)
 
  • #3
Nathanael said:
The link to the picture doesn't work (you need an account on that website)

This is why the forum rules say to attach the picture to the post and not to host it externally...
 
  • #4
Here's a general way to find net forces for non parallel/perpendicular forces.

##R^2 = R_{x}^2 + R_{y}^2##

##R_{x} = R\text{cos}(\theta)##

##R_{y} = R\text{sin}(\theta)##

For 2D motion:

##\vec{R} = <R_{x},R_{y}>##

if you decompose all your forces down into this vector form, then you can add all your vectors together simply by adding the x components and y components.

There's numerous trigonometric relationships that can be used to solve for your componts, and it's really as simple as using trig.
 

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  • #5
@ BiGyElLoWhAt
does R= F?
 
  • #6
would it be 10.2 cos 180?
 
  • #7
syavian1019 said:
does R= F?
R is an arbitrary vector, it can be Force or anything else (so yes)

Edit:
Nevermind, I didn't notice you edited your original post

syavian1019 said:
The two forces F1 and F2 act on a 27.0-kg object on a frictionless tabletop. If F1=10.2 N and F2=16.0 N find the net force on the object and its acceleration for cases (a) and (b).

(a) is 90 degrees, in the 3rd quadrant
(b) is 120 degrees, from the positive y-axis and moving clockwise

This is unclear.


You can't draw a picture or something?
 
  • #8
R is a general vector of length |R| and direction theta.

F can be equal to R, if you're seeking to break F down into components. If so, use the trigonometric relations I posted.

And as for 10.2 cos(180)

I have no clue, as I haven't seen the picture. I'll let you be the judge of that. ;)
 
  • #9
What do you mean by 90 degrees in the third quadrant?

Assuming you're labeling your quadrants as so:

II I
III IV

do you mean <-
or
|
v
?

(those are arrows btw haha)
 
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