What is the mathematical formula to calculate the force required to
push a weight up a wall when the pushing motion is at an angle to the
wall?
For example, assume that you are standing in front of a wall with a 3
foot pole. On one end of the pole is a 10 pound weight. You lean the
pole end with the 10 pound weight on the wall at about head height.
You hold the other end of the pole in your hand at an angle of 45
degrees to the wall, so that your hand is about at your stomach.
Assume that a linear slide ball bearing assembly is mounted between
the wall and the 10 pound weight. The coefficient of friction on the
bearing is 0.003, as measured by the bearing manufacturer when the
linear slide ball bearing assembly is mounted horizontally.
What is the mathematical formula to calculate how much force, in
pounds, is required to push the 10 pound weight up the wall using the
45 degree pushing motion? |
Request for Question Clarification by
mathtalk-ga
on
28 Oct 2004 15:04 PDT
Hi, jm900-ga:
Strictly speaking to push the weight up a wall requires only a bit
more force than the weight's component parallel to the pole to get it
started moving, and then depending on the velocity attained & the
dynamic friction present, enough force to maintain that velocity (the
frictional force will depend on the component of the weight
perpendicular to the pole).
While I understand that you want the formula and not merely a numeric
answer, with presumably enough explanation that similar problems could
be solved in the future, I'd like to clarify that your Question is
actually about the force required (and not, for example, about the
work required).
If you like, I can throw in a discussion of both quantities (work =
force * distance).
regards, mathtalk-ga
|
Request for Question Clarification by
mathtalk-ga
on
28 Oct 2004 15:14 PDT
Another Clarification, please. You wrote that the "linear slide ball
bearing assembly is mounted between the wall and the 10 pound weight."
I'm having difficulty picturing what is involved here. Since you
earlier state that the weight is at one end of the pole, and that the
goal is to "push a weight up a wall," I first assumed that the weight
would be pushed from one end of the pole to the other.
However with the concept of the pole sliding up the wall, I now notice
that the weight is already at the end of the pole on the wall.
Presumably this is where the linear slide ball bearing assembly is
mounted.
So, if my revised picture is correct, you will push along the
direction of the pole (ie. at a 45 degree angle to the horizontal),
thereby lifting the weight while maintaining the angle of the pole as
both pole and weight move upward.
Sound like the right picture?
regards, mathtalk-ga
|
Clarification of Question by
jm900-ga
on
29 Oct 2004 12:53 PDT
Hello mathtalk-ga,
I believe in your comment that you understand the position of the
components in my question. To further clarify, I have drawn the
component positions below. The 'Wall' is the vertical wall. The word
'Bearing' is typed vertically to represent the linear bearing mounted
between the wall and the weight. The 'WT' is the actual weight. The
'Pole' is written diagonally to represent the pole at its 45 degree
angle-the pole's angle never changes.
WAll
WAll
WAll B
WAll E
WAll A WT
WAll R P
WAll I O
WAll N L
WAll G E (Your hand holds the pole here, pushing the pole upwards
WAll while maintaining the 45* pole angle. The pole travels
WAll at 45* degrees, while the weight is lifted up the wall
vertically by sliding along the bearing)
I think that you may be on the right track in your comments.
I'm looking for the mathematical equation to lift this 10 pound weight
using this method with the pole at different angles to the wall, such
as at a 45* angle, or a 30* angle. I understand that more force will
be required the higher the angle of the pole (the closer the pole gets
to being in a horizontal position).
How much force is required to get the weight moving? I don't care
about distance at this time.
Thanks,
|
Request for Question Clarification by
mathtalk-ga
on
30 Oct 2004 07:39 PDT
Thanks, jm900-ga:
Please note the concise Comment provided by racecar-ga. The formula
there sensibly assumes that by "coefficient of friction" one means the
static coefficient of friction. There is also something known as a
"dynamic" or kinetic coefficient of friction, which models resistance
to motion after an object is in motion.
Let me know if your Question is already sufficiently addressed by the
Comment. The underlying idea is to resolve the force applied (along
angle T = 45 degrees) by the pole into a vertical force (upward) and a
horizontal force (against the wall). [Racecar-ga chose to define T as
the angle of the pole with respect to vertical, but of course when T =
45 degrees, the pole makes an equal angle with respect to horizontal.
So in your case cos(T) = sin(T) = SQRT(1/2).]
Some information on static vs. kinetic friction coefficients can be found here:
[Friction -- HyperPhysics/Georgia State Univ.]
http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html
Friction depends on the "normal force" (perpendicular to direction of
motion). The effect of friction here is slight (certainly compared to
neglecting the weight of the 3 foot pole), but racecar-ga has shown
how the coefficient of static friction appears in the formula for the
"threshold" force required to get the weight moving upward, so we can
see exactly how small the effect is.
regards, mathtalk-ga
|
