Understanding Equilibrium of Forces in a Triangle

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Homework Help Overview

The discussion revolves around the question of whether an object can be in equilibrium when subjected to forces of 10N, 20N, and 40N. Participants are exploring the conditions under which these forces can balance each other, particularly in the context of triangular force relationships.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are considering the vector sum of the forces and whether the combined effect of the 10N and 20N forces can counteract the 40N force. There is speculation about the arrangement of forces and their angles, as well as the implications of resolving forces into components.

Discussion Status

The discussion is actively exploring different interpretations of the problem. Some participants have offered insights into the limitations of the forces involved, while others are questioning the assumptions about the angles and directions of the forces. There is no explicit consensus yet, but productive dialogue is occurring.

Contextual Notes

Participants are working under the assumption that the forces can be arranged in a triangle, which leads to discussions about the properties of triangles and vector addition. The constraints of the problem, such as the requirement for equilibrium and the specific magnitudes of the forces, are also being considered.

drunkenfool
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The question is this.

"Is it possible for an object to be in a state of equilibrium when forces of 10N, 20N and 40N act upon it?"

I'm thinking yes, because you can always have the 10 N and 20 N force acting in the same direction and the 40N acting on both at a slight angle, equalling 30. But I don't know if I'm right. This seems like a really simple question, yet I don't know how to answer it. Any help would be greatly appreciated.
 
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Think of it this way: In order to completely balance the 40 N force, what must the other two forces add up to? Is that possible?
 
drunkenfool said:
The question is this.
I'm thinking yes, because you can always have the 10 N and 20 N force acting in the same direction and the 40N acting on both at a slight angle, equalling 30.

Your speculation here is very thoughtful. But remember, if you apply the 40N force at an angle, then there will be some of it not in the direction of the first two forces, meaning you will need to offset one of those as well, which will mean...etc etc.

All forces can always be resolved into their orthagonal components, so to keep it simple, work on it there. Apply the forces only in either the x or y directions.
 
...

Ah, so I guess the answer would be no. As if I keep it simple, like making all of them act on the x direction, there is no way that the 10N and 20N force can balance out the 40N force. Is that correct?

Oh, and thanks to the both of you.
 
The maximum force that can be applied with the help of 10N and 20N in a specified direction is 30N. that is their actual sum when the angle between them is 0deg. so you can't get a force of 40N with the help of other two forces in the direction opposite to the direction at which 40N force is applied.


when is the sum of three forces 0? AB + BC + CA = 0 in vector notations if they form a triangle. sum of two sides of a triangle can never be greater than the third side. it is the tht you get three vctors which can be properly placed such that their sum is 0.
 

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