Solve Force Problem with Four Masses in a Rectangle | Figure 12-23

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Homework Help Overview

The problem involves four masses positioned at the corners of a rectangle, requiring the calculation of the net force acting on one of the masses. The discussion includes the effects of changing the dimensions of the rectangle on the net force.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the positioning of masses and the forces acting on them. There are attempts to express the net force in vector form and questions about the nature of the forces involved, such as whether gravitational force should be considered. Some participants also inquire about the method for calculating distances between the masses.

Discussion Status

The discussion is ongoing, with some participants providing mathematical expressions for the forces while others express confusion about the problem's setup and the calculations involved. There is no clear consensus, but various interpretations and approaches are being explored.

Contextual Notes

Participants note the lack of clarity regarding the forces acting on the masses and the specific details of the problem's context, such as the gravitational force and distance calculations.

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Four masses are positioned at the corners of a rectangle, as indicated in Figure 12-23 (not to scale).

Figure 12-23 (attachment)

(a) Find the magnitude and direction of the net force acting on the 2.0 kg mass if x = 0.20 m and y = 0.15 m.

(b) How do your answers to part (a) change (if at all) if all sides of the rectangle are doubled in length?

I don't even know where to start on this problem, any help would be greatly appreciated.
 

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Hi wr1015.
Imagine a rectangle ABCD in counterclockwise direction, from botton to top. At vertice A there is mass m1, at B, m2, at C, m4 and at D m3.
I'm putting the origin of coordinate axis at m1. The vector force F on m1 is:
X -> x-axis versor
Y -> y-axis versor
x -> rectangle basis
y -> rectangle height

F=m1m2/x^2 X + m1m3/y^2 Y + m1m4/(x^2+y^2)(x X+y Y)/(x^2+y^2)^1/2

this is the net force done on m1 by all others including their direction. Grouping:

F=( m1m2/x^2 + m1m4x/(x^2+y^2)^3/2 ) X +
+ ( m1m3/y^2 + m1m4y/(x^2+y^2)^3/2 ) Y

This is the vector force. When you substitute your values, try to see if you can write the module of the vector times one linear combination of versors, so the direction apears explicity. Depending on the values, the direction can coincide with the diagonal of the rectangle.

In part (b), just take x->2x and y->2y

I hope this help you.
Cheers
 
Last edited:
I was wondering in what physics book was this problem located?


wr1015 said:
Four masses are positioned at the corners of a rectangle, as indicated in Figure 12-23 (not to scale).

Figure 12-23 (attachment)

(a) Find the magnitude and direction of the net force acting on the 2.0 kg mass if x = 0.20 m and y = 0.15 m.

(b) How do your answers to part (a) change (if at all) if all sides of the rectangle are doubled in length?

I don't even know where to start on this problem, any help would be greatly appreciated.
 
Im still confused about how to solve this problem. Isnt the force gravitational force, so you multiply by G on top? Also, how is that you find the distance between 1 and 4?
 

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