help with an inclined plane problem
Discussion
Spent too long on this and just cant figure it out. Plenty of stuff on the net about simple inclined plane problems but not quite like this.
In this problem the blue wedge can only move horizontally and the orange pin (viewed here end on) can only move vertically. It doesn't roll. The green wall is what is constraining the pin to move only vertically. It is fixed.
The idea of this mechanism (if you haven't already guessed) is that the pin is raised by pushing the wedge under it. The pin is pressed down by force Fdown. The wedge is moved horizontally by Fmove. Ignore gravity. Accounting for friction I need to determine the angle theta in order that the magnitude of Fmove <= Fdown.
As I say, lots of stuff on the net on inclined planes, friction / frictionless, accelerating / equilibrium, but none quite the same as this where the inclined plane is moving under the pin, which in turn is constrained by the wall.
Thank you please!
In this problem the blue wedge can only move horizontally and the orange pin (viewed here end on) can only move vertically. It doesn't roll. The green wall is what is constraining the pin to move only vertically. It is fixed.
The idea of this mechanism (if you haven't already guessed) is that the pin is raised by pushing the wedge under it. The pin is pressed down by force Fdown. The wedge is moved horizontally by Fmove. Ignore gravity. Accounting for friction I need to determine the angle theta in order that the magnitude of Fmove <= Fdown.
As I say, lots of stuff on the net on inclined planes, friction / frictionless, accelerating / equilibrium, but none quite the same as this where the inclined plane is moving under the pin, which in turn is constrained by the wall.
Thank you please!
nitrodave said:
45 degrees if you ignore friction and gravity, surely?
or am i missing something?
Yes, the bit where I said "accounting for friction" or am i missing something?
SpeckledJim said:
is the coefficient of friction between the wedge and the pin, and the wall and the pin, the same?
Lets say yes for now.
ok. Imagine this problem is a ball being rolled up a hill, you are the green line and it is you that is preventing the ball from rolling down. Now, do you see that the only internal angle that means the ball cannot be rolled is 90deg?
In your problem, as the pin is forced upwards, it must either roll up the inclined plane, which means something has to lift its weight off the green line, or it must slide upwards along the vertical green line and along the wedge to the right. You are asking the wedge to lift the pin from below but through an angle.
If you want to ignore gravity, and so the pin has no mass, then the force to lift the pin is zero.
In your problem, as the pin is forced upwards, it must either roll up the inclined plane, which means something has to lift its weight off the green line, or it must slide upwards along the vertical green line and along the wedge to the right. You are asking the wedge to lift the pin from below but through an angle.
If you want to ignore gravity, and so the pin has no mass, then the force to lift the pin is zero.
Sorry, yes I meant weight, obviously it has mass. However, ignoring gravity the force is still zero. Including gravity and answering the original question, there is no angle possible where Fmove <= Fdown. Even if friction between the pin and the green line was zero the wedge would still have to exert it's upward force through an angle and so Fmove > Fdown.
Sorry, yes I meant weight, obviously it has mass. However, ignoring gravity the force is still zero. Including gravity and answering the original question, there is no angle possible where Fmove <= Fdown. Even if friction between the pin and the green line was zero the wedge would still have to exert it's upward force through an angle and so Fmove > Fdown.
SpeckledJim said:
Without gravity, the pin won't have weight, but it'll still have mass.
To go from stationary to moving will require a force.
I don't think there's enough information in the question to answer it.
What do you think is missing? If something is missing then maybe that is why I'm struggling. I'm not sure that there is though!To go from stationary to moving will require a force.
I don't think there's enough information in the question to answer it.
longone said:
Sorry, yes I meant weight, obviously it has mass. However, ignoring gravity the force is still zero. Including gravity and answering the original question, there is no angle possible where Fmove <= Fdown. Even if friction between the pin and the green line was zero the wedge would still have to exert it's upward force through an angle and so Fmove > Fdown.
My instinct was that the inclined plane creates a mechanical advantage. If the system was frictionless then anything <45deg would mean Fmove < Fdown and it would be simple. With friction however the "breakeven" angle will be somewhat less that 45deg.I have been trying to construct free body diagrams for all the objects, forces and their reactions. That's what I'm struggling with. Maybe you're right and it is the case that Fmove can never be <= Fdown. If it is then I'd like to understand it and be able to construct the FBD showing it. I believe it is possible though.. Bare in mind that friction coefficients can vary too.
BTW just to make it clear, Fdown is some value >0. Masses and gravity are unimportant because they'd only be used to determine the force.. which as I've already said is Fdown and is >0.
Home that makes things clearer
Use Fdown = 30N if you like. Or say the mass of the pin is 3.058kg if you prefer.
Home that makes things clearer
Use Fdown = 30N if you like. Or say the mass of the pin is 3.058kg if you prefer.
SpeckledJim said:
FMove will increase the friction between the pin and the wall (by an unknowable amount). That's the bit I'm struggling with.
Will the pin rotate as it moves? Either gripping the wall or the wedge? Or is it non-rotating?
Me too!Will the pin rotate as it moves? Either gripping the wall or the wedge? Or is it non-rotating?
I think it's simpler to say the pin does not rotate. In real life it will of course, since despite the friction coefficients being the same, the normal forces will differ.
Edited by DudleySquires on Thursday 6th November 21:48
The inclined plane creates a mechanical disadvantage for lifting. It allows horizontal movement to provide some upward movement but only at the cost of Fmove always being greater than Fdown. The wedge is a mechanical transducer in its effect. It converts some of Fmove into Fdown, the proportion being related to the angle.
SpeckledJim said:
FMove will increase the friction between the pin and the wall (by an unknowable amount). That's the bit I'm struggling with.
Will the pin rotate as it moves? Either gripping the wall or the wedge? Or is it non-rotating?
Yes it will increase the friction between the pin and the wall, but it will also increase the friction between the wedge and the invisible wall the OP has not annotated. The wedge has to react somewhere. There is another 'green line' below the wedge??Will the pin rotate as it moves? Either gripping the wall or the wedge? Or is it non-rotating?
DudleySquires said:
Me too!
I think it's simpler to say the pin does not rotate. In real life it will of course, since despite the friction coefficients being the same, the normal forces will differ.
In real life the pin would not rotate in any practical situation where the included angle was greater than a few degrees, ie 0-3. That's how a wedge works. In real life you would have to shock the pin up by blows on the heel of the wedge.I think it's simpler to say the pin does not rotate. In real life it will of course, since despite the friction coefficients being the same, the normal forces will differ.
Edited by DudleySquires on Thursday 6th November 21:48
Thanks.
Here's where I've got to. I've got as far as balancing the pin, then the wedge, and now I'm considering how the friction & perp forces between wall and pin should be accounted for.
I've used a friction coefficient of 0.3, which has resulted in the "slip" beginning at ~17deg. When that happens, the wall begins to react some of the force, which also sets up a new perp and associated friction force between the wall and pin. I suppose that any perp force that cannot be reacted by the friction force will have to be reacted by the ground, which is in addition to the vertical force that is already present.
Here's where I've got to. I've got as far as balancing the pin, then the wedge, and now I'm considering how the friction & perp forces between wall and pin should be accounted for.
I've used a friction coefficient of 0.3, which has resulted in the "slip" beginning at ~17deg. When that happens, the wall begins to react some of the force, which also sets up a new perp and associated friction force between the wall and pin. I suppose that any perp force that cannot be reacted by the friction force will have to be reacted by the ground, which is in addition to the vertical force that is already present.
Edited by DudleySquires on Friday 7th November 12:12
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