Calculating Reaction Force: Box on Rigid Wall with Applied Force of 100 N

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Discussion Overview

The discussion revolves around calculating the reaction force exerted by a wall on a box when a horizontal force of 100 N is applied to the box. Participants explore the implications of friction and equilibrium in this scenario, considering various factors such as mass distribution and the coefficient of friction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes a scenario where a box of mass "M" is pushed against a rigid wall with a force of 100 N, questioning the reaction force from the wall.
  • Another participant notes that the solution may vary based on the geometry of the setup and the distribution of mass in the box.
  • A participant emphasizes that since the box is in equilibrium, the forces acting on it must balance, leading to the equation F_{1}=F_{f}+F_{2}, where F_{1} is the applied force and F_{2} is the wall's reaction force.
  • It is pointed out that the frictional force is defined as μN, where N is the normal force, which corresponds to the weight of the box, and that F_{1} is 100 N.
  • Another participant clarifies that μN represents the maximum static friction, and the actual frictional force can vary from 0 to this maximum, indicating that more information is needed for a definitive answer.

Areas of Agreement / Disagreement

Participants express varying views on the calculation of the reaction force, with some emphasizing the need for additional information to arrive at a single answer. There is no consensus on a definitive approach or solution.

Contextual Notes

Participants note limitations in the information provided, particularly regarding the geometry of the setup and the distribution of mass, which affect the calculation of the reaction force.

spectrum123
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Imagine a box of mass "M" lying on a surface and attached to a rigid wall too. Assuming that the co-efficient of friction between them(box and surface of ground) is "μ". and I am applying a force of 100 N horizontally on the box. So my question is how mush force is being applied by the wall on the box as a reaction force.
 
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You push towards the wall? There are multiple solutions, and it will depend on the geometry of your setup (where do you apply the force, how does the mass distribution of the box look like, and so on).
 


spectrum123 said:
Imagine a box of mass "M" lying on a surface and attached to a rigid wall too. Assuming that the co-efficient of friction between them(box and surface of ground) is "μ". and I am applying a force of 100 N horizontally on the box. So my question is how mush force is being applied by the wall on the box as a reaction force.

I don't quite follow you, since in the end you are not so interested in the friction but rather the force the wall exerts?

Anyway, I will try to answer with what I understand. Please clarify if this is not what you meant. The box is not moving, so it is in equilibrium. Considering all the forces in the horizontal direction, you get:

F_{1}=F_{f}+F_{2}
where F_{1} is the force you exert and F_{2} is the force the wall exerts. Note: Above F only concerns the magnitude, so we are ignoring direction for now.

You know Friction is μN and N has the same value as the weight (not mass) of the box. And F1 is 100N. So you just simply subtract the magnitude of the friction from 100N. That's the value of the force exerted by the wall.
 


Byron Chen said:
You know Friction is μN and N has the same value as the weight (not mass) of the box.
Realize that μN is the maximum possible value of static friction between the surfaces. The actual value can range from 0 to that maximum. As mfb says, there is not enough information to get a single answer.
 


Doc Al said:
Realize that μN is the maximum possible value of static friction between the surfaces. The actual value can range from 0 to that maximum. As mfb says, there is not enough information to get a single answer.

Good point I almost forgot. Guess I'm too used to problems dealing with kinetic friction.
 

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