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

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The discussion centers on calculating the reaction force exerted by a rigid wall on a box being pushed with a horizontal force of 100 N. The box is in equilibrium, meaning the forces acting on it must balance. The frictional force, determined by the coefficient of friction (μ) and the normal force (N), plays a crucial role in this calculation. The actual frictional force can vary from zero to its maximum value, which complicates the determination of the wall's reaction force. Ultimately, without specific details on mass distribution and friction, a definitive answer cannot be provided.
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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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