Forces and acceleration on frictionless table?

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Homework Help Overview

The discussion revolves around the forces acting on a 27.0-kg object placed on a frictionless tabletop, specifically focusing on two forces, F1 and F2, with given magnitudes and angles. The problem involves calculating the net force and acceleration for two different cases based on the angles provided.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method for calculating net forces in two-dimensional motion, including decomposing forces into their x and y components. There are questions about the correct formulas to use and how to interpret the angles associated with the forces.

Discussion Status

Some participants have offered general methods for finding net forces and decomposing vectors, while others express confusion about the angles and the setup of the problem. There is an ongoing exploration of how to visualize the forces and their directions, with no clear consensus reached yet.

Contextual Notes

Participants note issues with external links to diagrams, which may hinder understanding of the problem setup. There is also a lack of clarity regarding the angles and quadrants referenced in the original post.

syavian1019
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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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The link to the picture doesn't work (you need an account on that website)
 
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...
 
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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@ BiGyElLoWhAt
does R= F?
 
would it be 10.2 cos 180?
 
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?
 
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. ;)
 
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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