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

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The discussion focuses on calculating the net force acting on a 2.0 kg mass positioned at the corner of a rectangle with given dimensions. The net force is derived from the gravitational interactions between the masses at the rectangle's corners, using vector components for clarity. In part (b), it is noted that doubling the rectangle's dimensions affects the distances used in the force calculations but does not change the fundamental approach. Participants express confusion about the gravitational force and the method for determining distances between the masses. The conversation emphasizes the importance of understanding vector forces and their directions in solving the problem.
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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
 
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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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