What is the minimum force required to move the log?

Let A and B are end points of the log. Force F_h is applied in the horizontal direction to point A, so that the log starts barely accelerating. The log rotates around some point O. Since there are no force momenta in the direction along the log, point O lies on the log axis. The angular momentum balance is (neglecting small acceleration):

(1) F_h |OA| = μ (W-F_z) ( |OA|^2 + |OB|^2)/ (2L).

Here, F_z is the vertical component of the force. Eq. (1) is applicable if F_z is uniformly distributed along the log’s length. To find F_h, we minimize the expression (|OA|^2 + |OB|^2)/|OA|. This gives |OA|=L/√2 and

(2) F_h = α (W – F_z), α = (√2-1) μ = 0.083.

The total force is

(3) F^2 = F_z^2 + α^2 (W-F_z)^2.

Minimum of F is achieved at F_z = α^2 W/(1+α^2) and it is equal to

F_min = α W/√(1+α^2) = 0.08256 W = 82.56 N.

*******

For a rigorous treatment you need to go to elasticity theory.

A better approximation might be to assume a linear distribution of the reaction force along the log, which is

dN/dx = (W + 2 F_z)/L – 6 F_z x/L^2.

Here, x is the distance from point B. Then ∫ dN = W – F_z and ∫ x dN = WL/2 – F_z L, so that the vertical force and angular momentum balances are satisfied. The new horizontal angular momentum balance is

F_h |OA| = μ W ( |OA|^2 + |OB|^2)/2L – μ F_z (2 |OA|^2/L – |OA| (|OA|^2+|OB|^2)/L^2).

Minimizing this expression, we obtain a cubic equation for |OA|/L. In principle, this can be solved, and then substituted into the expression for the total force, which we again can minimize with respect to F_z, at least numerically. Such a solution looks ugly, and still it is based on a model assumption. It is not clear, whether it has some value.

One needs to find the minimum of

f^2 = f_h^2 + f_z^2,

provided

f_h = (μ/2) (2y + 1/y – 2) – μ f_z (4y – 1 – 2 y^2).

Here, f_h = F_h/W, f_z=F_z/W are normalized forces, and y = |OA|/L. Since analytics is cumbersome, I run a code, which gave:

f_min = 0.0817, achieved at f_z = 0.0134 and y = 0.7126.

Thus, F_min = 81.7 [N] in case of the linear distribution of the reaction force.

I’ll just try my guess here.

…………………FsinØ

……………………..↑………..

……………………..│…….⁄.F

……………………..│….⁄…Ø

……………………..│.⁄–→FcosØ

…..……..….__—‾‾‾‾‾‾‾—-__

..…….._-‾‾……………..,…..‾‾-_

………/….…………………..….‾-_

….…/….…………………..…..…

……|………………+———r——→|

……………………|….…….……./

.…………………..|d….….……./

………‾-_………….|…………._-‾

…………‾–__…….|..……_–‾

…..….……….‾‾‾‾–●—‾‾‾‾

Point of pivot……B

My interpretation of the log not being able to roll is that it will pivot at B, even just for a nanosecond (virtual motion) because on a horizontal surface it has nowhere to go, except for that little “dent” in the surface.

Torque is given as

T = Іα

……….Moment of Inertia I = ½mr ² + md²

……….d = new gyration radius (from Parallel-axis theorem)

………..d = r …<=====(centroid to new axis)

FcosØ(2r) =[½mr ² + md²]α

FcosØ(2r) =[½mr ² + mr²]α

FcosØ(2r) =[½mr + mr]rα……..divide by r

FcosØ(2) =[½mr + mr]α

………………=3/2mrα

F = 3 mrα / 4cosØ……OR

F =(3/4)mrα secØ

(So if it is specified how fast it will be accelerated, an equivalent F could be found)

(Just wondering, why do we need μ since the log will not be made to roll?)

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Edit: (as per clarification from Dr D)

..FsinØ

…..↑………….F

…..│……….⁄.

…..│…….⁄………..▬►motion

…..│….⁄…Ø

W↓│.⁄———→FcosØ

….☻←—Ff

▒▒↑▒▒▒▒▒▒▒▒▒▒

…..N

The net unbalanced force = FcosØ – Ff

FcosØ – Ff = ma

…………….. Ff = µN

……………….N = W – FsinØ

……………….Ff =µ(W – FsinØ)

…………………..= µ(mg – FsinØ)

FcosØ – µ(mg -FsinØ) = ma

FcosØ – µmg +µFsinØ = ma

FcosØ +µFsinØ) = µmg +ma

F(cosØ +µsinØ) = m(µg +a)

F = m(µg+a) / (cosØ +µsinØ)

(It seems, my answer never included the spinning like the Helicopter Blades. I missed the question entirely)

Thanks, there’s time yet for another solution.

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The following solution may just be conceptual.

..axis of rotation——→|

…………….….…._ —-|.._

…………….……⁄…………….

………….…..⁄…………….…|B

………Y…⁄…….……..…….⁄

….……|.⁄…/F……….…….⁄◄▬Ff

……..⁄..|…/..….⁄………..⁄

… .⁄_–‾|‾‾/– .⁄_…….. .⁄

./……..|./ .⁄…….. .⁄

|……..+———-|.⁄——-X

._… .⁄…….._/

…‾-.⁄-__|_—‾

Z.⁄…….A

The application of the force F might as well be on the center of mass (as per additional details)Assuming the log as having a point cross sectional area

….…Fy.

…..….↑…./F

‘’……..|…/..….……..

.. ._–‾‾|‾./– ._..

./……..|./Ø……. .

|……..+———|—–→Fx

._… .⁄…….._/ ◄▬Ff

…‾-.⁄-__|_—‾

Z.⁄…….A

View ┴ to X-Y plane

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

.Fy……….…W

..↑←–½L—→↓

A▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀B

.↑↑↑↑↑↑↑↑↑↑↑↑ ▲↑↑↑↑↑↑↑N(max)

N(min)………….▐ ←⅓L→|.

…………….………N

View ┴ to Z -Y plane

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

It should be considered that there is no distinct point of application for any Reactive force supporting the log from the ground. Only the resultant N which we have to assume must be acting at (⅓L) from point (B) because the distribution of pressure under the log is such that it is minimal at end A and maximum at end B. (This may be seen when, for instance, the ground where this log is placed happens to be soft, the log will sink deeper at point B.

.……….…

..|←–½L—→|

A▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀B← point of pivot

. ↓………………. ▲

.Fx……………….▐ ←⅓L→|.

…………..………Ff

View ┴ to Z – X plane

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

From the above reasoning, the point of greatest resistance to rotation will be at point B, and hence, the log will pivot at this point.

Torque generated by Fx will rotate the log, giving it angular acceleration.

T = Іα

Fx(L) – Ff(⅓L) = Іα

FcosØL – µmg(⅓L)= ⅓mL² α….divide by L

FcosØ -⅓µmg = ⅓mLα

F = ⅓m(µg + αL)/cosØ

F = m(µg + αL) / 3cosØ

=========================

The minimum force that will just move the log

∑Mв = 0

0 = – Fx (L) + Ff (⅓L)

Fx(L) = µmg(⅓L)…………divide by L

FcosØ = ⅓ µmg

F = µmg /3cosØ

1

The reason engineers invented strain gauges was to deal with excessively brainy physicists and their hypotheticals. (just kidding)

I’ll get to this tonight if it is still open.