| Topic: |
Science > Physics |
| User: |
"Nikyu" |
| Date: |
18 Apr 2006 07:51:27 AM |
| Object: |
Static Equilibrium? |
It's been a while that I've had to deal with a statics problem but
I came across this one and am having trouble figuring it out. I have
what is basically a spring loaded plunger. One end is pinned to the
floor the other is leaning against a wall. So there is some spring
force trying to push the plunger out, however, because of the static
friction on the wall the plunger does not slide against the wall. So
do I include the internal spring force? If so, how?
Second problem is what if the spring force is great enough that the
plunger can overcome static friction and begin to move at a constant
velocity. So now the plunger is sliding up the wall and rotating at
its pin at its base. Now forget the fact that the spring will be
extending and loosing force and that the angle for the reaction forces
changes...just assume the movement is small enough to neglect that
stuff. I believe this is still a static problem because there are no
accelerations. But is this basically the same problem as in the
previous case?
My guess is that since one end can move freely, it does not matter how
much force is in the spring...as long as there are no accelerations,
just the plunger weight, friction and reaction forces matter. But
I'd like to hear others opinions.
Thanks,
Eric
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| User: "PD" |
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| Title: Re: Static Equilibrium? |
18 Apr 2006 08:44:02 AM |
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Nikyu wrote:
It's been a while that I've had to deal with a statics problem but
I came across this one and am having trouble figuring it out. I have
what is basically a spring loaded plunger. One end is pinned to the
floor the other is leaning against a wall. So there is some spring
force trying to push the plunger out, however, because of the static
friction on the wall the plunger does not slide against the wall. So
do I include the internal spring force? If so, how?
Second problem is what if the spring force is great enough that the
plunger can overcome static friction and begin to move at a constant
velocity. So now the plunger is sliding up the wall and rotating at
its pin at its base. Now forget the fact that the spring will be
extending and loosing force and that the angle for the reaction forces
changes...just assume the movement is small enough to neglect that
stuff. I believe this is still a static problem because there are no
accelerations. But is this basically the same problem as in the
previous case?
My guess is that since one end can move freely, it does not matter how
much force is in the spring...as long as there are no accelerations,
just the plunger weight, friction and reaction forces matter. But
I'd like to hear others opinions.
Thanks,
Eric
This is a pretty basic problem if you start drawing it, which I
recommend.
The force of the spring F is pointing at an angle. It therefore has a
component Fy which is along the surface of the wall and a component Fx
that is perpendicular to the wall. The ratio between Fx and Fy is
determined by the angle of the spring, obviously. There will be no
slippage as long as the friction can be as great as Fy. The limit of
the friction, on the other hand, is u*Fx, where u is the coefficient of
static friction. There is obviously then some connection between the
angle of the spring and the coefficient u. Do you see what it is?
PD
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| User: "Nikyu" |
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| Title: Re: Static Equilibrium? |
18 Apr 2006 02:22:45 PM |
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I understand how the friction force will change with the angle. But I
really cannot seem do draw the correct diagram. Let me explain. Take
the example of a ladder leaning against a wall and let's assume it is
pinned at the floor. Lets say the ladder is spring loaded so that it
pushes out...kind of how a firemans ladder extends out from the truck.
Now I can draw a free body diagram, do a force balance that includes
the weight and friction and the reactions at the wall and pin but that
does not take into account the internal spring force. So what I try to
do is cut the ladder in half. On the bottom half I have the pin
reaction forces, half of the weight acting at the center of my cut
ladder and the compressed spring force acting along the ladder pointing
toward the pin. On the top half of the ladder I have the wall reaction
force, half of the weight acting at the center of the cut ladder, the
wall friction and the spring force acting along the ladder pointing to
the pin. But my problem or confusion is that I am separating the wall
reaction forces from the pin reaction forces by cutting the ladder but
don't the wall reaction forces contribute to the pin reaction forces?
Eric
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| User: "Nikyu" |
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| Title: Re: Static Equilibrium? |
18 Apr 2006 02:35:55 PM |
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Let me add one more thing. If I cut the ladder in half and do a free
body diagram of the upper half of the ladder then I don't know what
the reaction at the wall is from the weight because to determine that I
would take the moments about the pin which is not included in the upper
half. Make sense?
If I want to determine the reaction (normal force) from the wall due to
the weight only, I would take the moment about the pin...no problem.
Now include the spring force and it adds to the reaction force at the
wall. So do I just simply take the cosine of the angle and get the
component of the spring force that pushes against the wall and add it
to what I already have determined for weight reaction force?
I'm really confusing myself!
Eric
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| User: "PD" |
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| Title: Re: Static Equilibrium? |
18 Apr 2006 02:53:22 PM |
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Nikyu wrote:
I understand how the friction force will change with the angle. But I
really cannot seem do draw the correct diagram. Let me explain. Take
the example of a ladder leaning against a wall and let's assume it is
pinned at the floor. Lets say the ladder is spring loaded so that it
pushes out...kind of how a firemans ladder extends out from the truck.
Now I can draw a free body diagram, do a force balance that includes
the weight and friction and the reactions at the wall and pin but that
does not take into account the internal spring force. So what I try to
do is cut the ladder in half. On the bottom half I have the pin
reaction forces, half of the weight acting at the center of my cut
ladder and the compressed spring force acting along the ladder pointing
toward the pin. On the top half of the ladder I have the wall reaction
force, half of the weight acting at the center of the cut ladder, the
wall friction and the spring force acting along the ladder pointing to
the pin. But my problem or confusion is that I am separating the wall
reaction forces from the pin reaction forces by cutting the ladder but
don't the wall reaction forces contribute to the pin reaction forces?
Eric
Be sure to define the free-body that you are drawing a free-body
diagram for.
If you are using the ladder for the free body, then the spring-force is
not a force on that body, so it won't appear as the forces acting on
*that* body.
Here's a hint: Because of the presence of the spring force, you're
going to need to draw the free-body diagram for *three* bodies: The
ladder/spring, the pin at the floor, and the wall. For example, the pin
at the floor feels the force of the ladder pushing on it due to the
spring force, and this force is numerically equal to the force of the
ladder pushing on the wall in another free-body diagram.
PD
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| User: "Nikyu" |
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| Title: Re: Static Equilibrium? |
18 Apr 2006 03:42:48 PM |
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Yes, that makes sense. It's just like how a 4 bar mechanism is dealt
with. I'll try that approach.
Thanks!
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