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- Thread starter spectrum123
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- #1

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- #2

mfb

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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).

- #3

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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[itex]_{1}[/itex]=F[itex]_{f}[/itex]+F[itex]_{2}[/itex]

where F[itex]_{1}[/itex] is the force you exert and F[itex]_{2}[/itex] 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.

- #4

Doc Al

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Realize that μN is theYou know Friction is μN and N has the same value as the weight (not mass) of the box.

- #5

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Realize that μN is themaximumpossible 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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