Can 3 forces of 9N, 4N, and 6N be in equilibrium?

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SUMMARY

The discussion centers on the equilibrium of a mass subjected to three forces: 4.0 N, 6.0 N, and 9.0 N. It is established that these forces can be in equilibrium if the angles between them are appropriately chosen, allowing the resultant of the 4 N and 6 N forces to equal 9 N. When the 9 N force is removed, the new net force can be calculated, leading to the determination of the mass's acceleration using Newton's second law, F=ma. The key takeaway is that the system's equilibrium depends on the vector sum of the forces and their directions.

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  • Knowledge of vector addition and resultant forces.
  • Familiarity with the concept of equilibrium in physics.
  • Basic skills in drawing force diagrams and applying the law of cosines.
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MattDutra123
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Homework Statement


A mass of 3kg is acted upon by three forces of 4.0 N, 6.0N, and 9.0N and is in equilibrium. The 9N force is suddenly removed. Determine the acceleration of the mass.

Homework Equations


F=ma.

The Attempt at a Solution


My main problem with this question is that I cannot think of any situation in which these three forces would be in equilibrium. The question does not specify in which direction the forces are in or any angles (in the case of an inclined plane). How can these forces possibly be in equilibrium?
 
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Welcome to the PF.

Well, 4+6=10 > 9, so it should be possible. Just think of ways that the sum of the x-components and the sum of the y-components can both be zero. Draw a diagram and write the two equations, and see if you can solve for the relevant angles... Please show us your sketch (use the UPLOAD button) and write out your equations. Thank you. :smile:
 
You could think of it in terms of the 4 and the 6 acting together to cancel out the 9. Is that possible?

Suppose the 4 N and the 6 N act together in the same direction. What is their resultant force? What happens if you use that force to oppose the 9 N force?
Suppose the 4 N and the 6 N act in directly opposite directions. What is their resultant force? What happens if you use that force to oppose the 9 N force?
If the 4 N and 6 N are at an angle relative to each other, the situation will be in between those two situations.
 
You can the law of cosines to compute what angle the 4N and 6N need to be relative to each other to make a resultant of 9N.
 
Charles Link said:
You can the law of cosines to compute what angle the 4N and 6N need to be relative to each other to make a resultant of 9N.
Just to point out that this is not necessary to solve the problem.
MattDutra123 said:
The question does not specify in which direction the forces are in or any angles (in the case of an inclined plane).
The main thing, as has been pointed out is that the forces can be in equilibrium given appropriate angles between them. What does this tell you about the resultant of the 4 N and 6 N forces?
 
Are you familiar with a "triangle of forces" in equilibrium? Can you draw a triangle with sides of those lengths?
 
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You are thinking too hard. If the system is in equilibrium, then a = 0. You also know F = 0 and F is the sum of the forces. So if you remove one force you should be able to calculate the new sum of forces and therefore the acceleration.
 
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I've never studied forces beyond high school, so I've never encountered problems like this but, unless I miss my guess, this seems to require nothing more than basic arithmetic.
 
I think @haruspex has the best answer for this problem in post 6.
 
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Eric Bretschneider said:
You are thinking too hard. If the system is in equilibrium, then a = 0. You also know F = 0 and F is the sum of the forces. So if you remove one force you should be able to calculate the new sum of forces and therefore the acceleration.
That is the neatest way to solve the given problem, yes, but the OP's question in the forum is a little different.
 

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