Mechanical force question on equilateral triangle

AI Thread Summary
The discussion revolves around a physics problem involving three particles at the corners of an equilateral triangle, where the goal is to determine the net force acting on one particle (m1) given that the resultant force in the x-direction is zero. Participants clarify that if the x-component of the forces acting on m1 is zero, the forces from the other two masses (m2 and m3) must be equal and opposite in the x-direction. The symmetry of the triangle suggests that m3 must equal m2 to maintain this balance, leading to the conclusion that m3 is 1.2 kg. The conversation emphasizes the importance of sketching force vectors to visualize the problem and correctly apply trigonometric functions to find components of the forces. Overall, the discussion highlights the principles of force balance and vector decomposition in physics.
joeycyoon
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Homework Statement


Three particles with masses m1=2.5kg, m2=1.2kg and unknown mass m3 are placed at the corners of an equilateral triangle of side 30.0cm. If the resultant force in x-direction acting on m1 is zero, find the net force acting on m1.


Homework Equations


F=(Gm1m2)/R^2
Fx=Fcosθ
Fy=Fsinθ
F=sqrt/(Fx)^2 +(Fy)^2

The Attempt at a Solution


Since Fx on m1=0
Fm1 cos 60 = 0
Fm1 = 0
The Fy on m1 = 0
Fy = Fm1 sin 60
Fy = 0 sin 60
Fy = 0
So does it mean the net force acting on m1 is
F=sqrt/(0)^2 +(0)^2
F=0N ?
 
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Is that the complete problem statement that you were given? Was there an image?

It seems to me that by simply rotating the triangle to orient the net force on m1 to lie along the y-axis (so no x-component) , m3 could take on any value at all and there would be in infinite number of solutions.
 
gneill said:
Is that the complete problem statement that you were given? Was there an image?

It seems to me that by simply rotating the triangle to orient the net force on m1 to lie along the y-axis (so no x-component) , m3 could take on any value at all and there would be in infinite number of solutions.

Emm the m1 on top of the triangle and on y-axis, m2 and m3 lie on x-axis, which m3 at the left and m2 at the right.
 

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joeycyoon said:
Emm the m1 on top of the triangle and on y-axis, m2 and m3 lie on x-axis, which m3 at the left and m2 at the right.

Okay, by symmetry you should be able to conclude something about the mass of m3: if there's to be no x-component in the net force on m1, what can you say about the magnitude of the two forces acting on it (one from each mass, m2 and m3)?

A good idea would be to sketch the force vectors on m1 along with the resultant of those forces. How does your sketch compare with your previous conclusion about the net force?
 
gneill said:
Okay, by symmetry you should be able to conclude something about the mass of m3: if there's to be no x-component in the net force on m1, what can you say about the magnitude of the two forces acting on it (one from each mass, m2 and m3)?

A good idea would be to sketch the force vectors on m1 along with the resultant of those forces. How does your sketch compare with your previous conclusion about the net force?

Is it means one of the magnitude acting on it is negative?
Seriously, I don't know how to sketch the force vectors :confused:
 
Between point masses the gravitational force acts along the line joining the points and is always an attraction (so the the force vector always points toward the mass that is responsible for the force).

Magnitudes are never negative. By definition they are an absolute value (thus positive). What can change is the direction that the vector points.

Vectors can be decomposed into components that are parallel to the coordinate axes. Note that unlike magnitudes, components can take on positive or negative values depending upon which direction they point along the given coordinate axis. You should have learned about this and how to sum the components to find a resultant vector when two or more forces are acting on a single "target". If you want the resultant vector to have a zero x-axis component, then the x-components of the acting forces must sum to zero.

So. Sketch your force vectors acting on m1 as arrows pointing from m1 to the individual masses m2 and m3. By symmetry you should be able to see how to "balance" the forces so that the net x-component is null.
 
gneill said:
Between point masses the gravitational force acts along the line joining the points and is always an attraction (so the the force vector always points toward the mass that is responsible for the force).

Magnitudes are never negative. By definition they are an absolute value (thus positive). What can change is the direction that the vector points.

Vectors can be decomposed into components that are parallel to the coordinate axes. Note that unlike magnitudes, components can take on positive or negative values depending upon which direction they point along the given coordinate axis. You should have learned about this and how to sum the components to find a resultant vector when two or more forces are acting on a single "target". If you want the resultant vector to have a zero x-axis component, then the x-components of the acting forces must sum to zero.

So. Sketch your force vectors acting on m1 as arrows pointing from m1 to the individual masses m2 and m3. By symmetry you should be able to see how to "balance" the forces so that the net x-component is null.

Is my force vectors correct? Then the components of x will be negative and positive, which will sum up to zero.
 

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joeycyoon said:
Is my force vectors correct? Then the components of x will be negative and positive, which will sum up to zero.

That's the idea. What value of mass m3 will make your diagram true?
 
gneill said:
That's the idea. What value of mass m3 will make your diagram true?

Its 1.2 kg!
One more to ask, to find the x-component which using cos theta, we use the theta as the value from positive x-axis right?
 
  • #10
joeycyoon said:
Its 1.2 kg!
One more to ask, to find the x-component which using cos theta, we use the theta as the value from positive x-axis right?

It depends on the orientation of triangle you're looking at. It's best to sketch in the angle you're going to use and determine its value from whatever reference direction is convenient. Assign the sign to the component by verifying its direction on your sketch.

Sometimes you're given a different angle to work with, such as the angle with respect to the vertical (y-axis), in which case you would use the sine of the angle in calculating the x-component!

attachment.php?attachmentid=73309&stc=1&d=1411133278.gif
 

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  • #11
gneill said:
It depends on the orientation of triangle you're looking at. It's best to sketch in the angle you're going to use and determine its value from whatever reference direction is convenient. Assign the sign to the component by verifying its direction on your sketch.

Sometimes you're given a different angle to work with, such as the angle with respect to the vertical (y-axis), in which case you would use the sine of the angle in calculating the x-component!

attachment.php?attachmentid=73309&stc=1&d=1411133278.gif


Get it clearly!
Thanks for the explanation through all these!
:thumbs:
 
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