Vector Mechanics Statics question

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Discussion Overview

The discussion revolves around a vector mechanics statics problem involving a wedge and a force P that is removed after insertion. Participants are exploring the implications of this removal on normal and friction forces, as well as the conditions for static equilibrium. The scope includes technical reasoning and conceptual clarification related to static friction and normal forces.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the scenario, questioning how to calculate forces after the removal of force P.
  • Another participant suggests that the removal of force P would create a net restoring force upwards on the wedge, which is balanced by static friction, indicating a need to calculate horizontal components of force.
  • Several participants emphasize the importance of determining the normal force during the insertion phase and caution against assuming the wedge remains stationary after P is removed.
  • There is a discussion about whether the normal force changes once P is removed, with differing views on the implications for static equilibrium.
  • One participant argues that the wedge can only remain in place if the force does not exceed static friction, while another insists that it should be assumed to remain in static equilibrium for the problem to be solvable.
  • The role of static friction is highlighted as crucial for preventing the wedge from moving, with some participants asserting that it is sufficient to keep the wedge in place under certain conditions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the normal force changes after the removal of force P, and there are competing views on the assumptions necessary for solving the problem. The discussion remains unresolved regarding the conditions under which the wedge remains stationary.

Contextual Notes

There are limitations regarding the assumptions made about the wedge's movement and the coefficients of friction, which are not provided in the problem statement. The discussion reflects uncertainty about the static and dynamic friction forces involved.

ashishsinghal
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In the question I am not able to understand what the scenario is. Is it that we are removing the P force after insertion? If that is the case, both the normal and the friction forces will change after removal of P. Then how can we calculate.
Please help, I am going to have my exams tomorrow.
 

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It reads that the downwards P force has been removed - then there would be a net restoring force upwards on the wedge equal to P which is balanced by the static friction.

You need to work out what the horizontal components the force has to be to make the friction force big enough to do that. Solving it requires understanding of how static friction works.
 
Last edited by a moderator:
There's more to it than that. First need to work out what the normal force is during the last little part of the insertion. Bear in mind that that force is normal to the cut surface, not normal to P. So be careful in calculating the (peak) force that P has to match.
Having determined that normal force, you can calculate the maximum frictional force that is available to prevent the wedge slipping back out, and how much force is needed to achieve that. Note that you should not assume the wedge stays put.
 
Last edited by a moderator:
haruspex said:
There's more to it than that. First need to work out what the normal force is during the last little part of the insertion. Bear in mind that that force is normal to the cut surface, not normal to P. So be careful in calculating the (peak) force that P has to match.
Having determined that normal force, you can calculate the maximum frictional force that is available to prevent the wedge slipping back out, and how much force is needed to achieve that. Note that you should not assume the wedge stays put.

But wouldn't the normal force change once P has been removed?
 
Note that you should not assume the wedge stays put.

Why not? : You have no information that it moves or about the coefficient of sliding friction.

You should assume that 25 lbs is just sufficient to place the wedge where you see it and that it remains there in static equilibrium.

Otherwise the problem cannot be solved with the information provided.
 
ashishsinghal said:
But wouldn't the normal force change once P has been removed?
I think you do have to assume that the split ring is fairly elastic, so the 'closing' force it exerts is the same, for a given degree of opening, whether being forced open or being allowed to close a little.
Studiot said:
You have no information that it moves or about the coefficient of sliding friction.
No, but it's the static friction that matters to prevent the ring starting to close again, and we do know that.
 
No, but it's the static friction that matters to prevent the ring starting to close again, and we do know that.

Ergo it doesn't move.
 
Studiot said:
Ergo it doesn't move.
It will move if the force tending to eject the wedge exceeds static friction. (Dynamic friction can never exceed static friction.) The wedge will stay put if the force is less than static friction. Ergo, static friction is what matters.
In the present case, it is easily demonstrated that the wedge will stay put, but it should not be assumed.
 

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