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In a recent publication the author derived and experimentally tested several theoretical models, distinguished by different boundary conditions at the contacts with horizontal and vertical supports, that predicted the forces of reaction on a fixed (
*i.e.* inextensible) ladder. This problem is statically indeterminate since there are 4 forces of reaction and only 3 equations of static equilibrium. The model that predicted the empirical reactions correctly used a law of static friction to complement the equations of static equilibrium. The present paper examines in greater theoretical and experimental detail the role of friction in accounting for the forces of reaction on a fixed ladder. The reported measurements confirm that forces parallel and normal to the support at the top of the ladder are linearly proportional with a constant coefficient of friction irrespective of the magnitude or location of the load, as assumed in the theoretical model. However, measurements of forces parallel and normal to the support at the base of the ladder are linearly proportional with coefficients that depend sensitively on the location (although not the magnitude) of the load. This paper accounts quantitatively for the different effects of friction at the top and base of the ladder under conditions of usual use whereby friction at the vertical support alone is insufficient to keep the ladder from sliding. A theoretical model is also proposed for the unusual circumstance in which friction at the vertical support can keep the ladder from sliding.

In a recent publication [

Despite its practical importance to workplace safety [

very few cases where the system included friction at the wall, the problem did not call for, nor provide, expressions for the reactions at static equilibrium [

In general, the solution to a statically indeterminate mechanics problem calls for complementary information ordinarily in the form of a restrictive boundary condition or supplemental external force. Examples of the former, commonly found in textbooks and treatises covering the mechanics of continuous or deformable media, are cases of a statically indeterminate axially or transversely loaded beam [

In part I the author systematized the different approaches into three distinct fundamental physical models on the basis of the constraints imposed at the ground and wall supports. Models 1 and 2 involved constraints on axial compression and bending deformation. In Model 3, the ladder was free to rotate (but not slide) about the contact point at the ground and to displace vertically against friction at the contact with the wall. The three models were solved analytically, and the predicted forces of reaction of each model were compared with measured forces of reaction as a function of load P and load location (as a fraction β of the ladder length L) measured from the origin at the ground. The experimental reaction forces, measured on both a fixed ladder and on an actual E-B beam, agreed well with predictions of Model 3 and disagreed markedly with predictions of the other models. The theoretical expressions for the reaction forces were therefore found to depend only on the ladder weight W, load P, angle of inclination θ with the ground, and coefficient of friction μ between the ladder or beam and the wall―and not on E, A, and I.

The phenomenological force of static friction adopted by the author in Model 3 was based on two principles often referred to as Amontons laws [

・ The maximum force of friction is directly proportional to the applied load. (1)

・ The force of friction is independent of the apparent area of contact. (2)

Application of Equation (1) to the contact at the wall related the vertical reaction R 4 to the horizontal reaction R 3 by a linear expression

R 4 = μ R 3 (3)

in which the coefficient μ is a constant depending only on the surfaces of the beam and wall, and not on the angle of inclination of the beam or the magnitude or location of impressed loads. Equation (3) removed the indeterminacy of the problem and, together with the 3 equations of static equilibrium, led uniquely to the solution [

R 3 = R 2 = R 4 μ = 1 2 W + β P tan θ + μ (4)

R 1 = W + P − μ R 3 = W ( tan θ + 1 2 μ ) + P ( tan θ + ( 1 − β ) μ ) tan θ + μ (5)

where the first equality in Equation (4) expresses the vanishing of the net horizontal force in static equilibrium, and the first equality in Equation (5) expresses the vanishing of the net vertical force.

The coefficient μ is not predicted by the model, but is obtained empirically from the data (i.e. the measured reactions) by a method of visual inspection and adjustment [

Agreement between theory (Model 3) and experiment notwithstanding, it is important for both conceptual and practical purposes to show explicitly how well the reactions at the wall satisfy the assumed law of static friction. This demonstration is given in Section 2.

Resolution of the ladder problem by complementary information in the form of the proportionality relation Equation (3) raises an important general question regarding the application of Amontons laws. It is to be understood, of course, that the laws of friction are not physical laws on par with such universal principles as the laws of thermodynamics or the law of conservation of electric charge. Rather, statement (1) is a kind of constitutive relation that depends on the properties of materials and the geometry of their surfaces. Nevertheless, the laws of friction find wide application throughout science and engineering, and it is necessary to be able to apply them correctly.

The question that arises is this: If the forces of reaction parallel and normal to the wall are related by Equation (3), can one likewise relate the forces of reaction parallel ( R 2 ) and normal ( R 1 ) to the ground by a comparable expression

[ R 2 = μ g R 1 ] (6)

where μ g is the corresponding frictional constant and can differ from μ ? The equation is enclosed in square brackets to emphasize that it is a conjecture to be examined, and not an established equality.

The short answer to the question is “No”. Equation (3) and Equation (6) cannot both be correct because these 2 independent relations, together with the 3 equations of static equilibrium, constitute 5 independent relations to determine 4 unknown forces. Thus, the system would be over-determined and there would in general be no self-consistent solution. Since the experiments in Part I provided strong evidence in support of Equation (3), it must follow that Equation (6) cannot be correct as it stands. But, given that there is friction at both the wall and ground, why is there asymmetry in the application of the law of static friction, and what is the theoretically valid relation that replaces Equation (6)? These questions are answered in Section 2.

In Section 2 the experiments in Part I are briefly recapitulated with additional technical details concerning the measurements of pressure, shear, and strain. The measured forces of reaction at the wall are shown to confirm directly the validity of Equation (3). An exact expression relating the parallel and normal forces of reaction at the ground is derived and tested against experimental results. Issues concerning the apparent asymmetric application of the laws of static friction are resolved.

Conclusions to this work are summarized in Section 3.

An analogous setup (not shown) was employed to measure the reactions on a single Euler-Bernoulli wood beam ( L b = 204 cm , W b = 1 5.3 lbs = 67.9 N ) inclined at θ b = 57.7 ∘ to the ground. Loads were applied in units of 2 kg masses stacked on a weight hanger of 444 g hooked to eyelets at positions β = ( 42 , 72 , 102 , 132 , 162 , 192 ) / 204 along the beam length.

External and internal features of the force platforms are respectively shown in frames A and B of

rectangle on the front face of the load cell in frame A. When a deformation stretches the metal foil, as shown (to an exaggerated extent for visibility) in the top strain gauge of frame B, the resistance of the foil is slightly increased. Correspondingly, when a deformation compresses the metal foil, as shown in the bottom strain gauge of frame B, the resistance of the foil decreases slightly. The degree of strain is proportional to the difference in resistance of the upper and lower strain gauges, as determined by means of a Wheatstone bridge circuit which detects a potential difference between the two foils. As a typical example, a bridge excitation voltage of 10 V can give rise to a detectable output voltage on the order of millivolts, thereby permitting measurement of strains (length change per initial length) of a few parts in 1000.

The relation between parallel and normal forces of reaction on a single Euler-Bernoulli wood beam is shown graphically in

The experimental points in panel B are consistent with the assumed law of static friction in which the parallel reaction at the wall is the same linear function of the normal reaction at the wall irrespective of the magnitude and location of the load. The superposed linear function (dashed black line), calculated theoretically from Equation (3) and Equation (4), bears out Amontons 1st law for a frictional coefficient μ ≈ 0.31 at the wall. Although there is a narrow spread of points about the dashed line, each set of 4 points of the same color approximates a line of the same slope―and therefore same value of μ ―as the theoretical line.

In contrast to the single linear relation in panel B, the pattern of experimental points in panel A shows unambiguously that each set of 4 points of a given color (i.e. variation of R 2 and R 1 with P for fixed load location β ) follows a linear relation whose slope and therefore frictional coefficient depend on β . The solid lines in panel A were calculated theoretically from Equation (4) and Equation (5) for the same wall coefficient μ ≈ 0.31 as in panel B. Thus, the conjectured Equation (6) must actually take the form

R 2 = μ g ( β ) R 1 . (7)

The functional form of μ g ( β ) can be derived from Equation (4) and Equation (5) by noting that a differential change in load dP at constant W and θ results in differential changes in reactions R 1 and R 2 of the form

d R 2 = K β d P d R 1 = K ( tan θ + μ ( 1 − β ) ) d P (8)

where K is a constant. It then follows that the slope

μ g ( β ) = d R 2 d R 1 = β tan θ + μ ( 1 − β ) (9)

depends on load location β , but is independent of the magnitude P of the load. Thus, as a consequence of the law of static friction (3) at the wall support, a modified linear force law, Equation (7) with slope given by Equation (9), turns out to be applicable at the ground support.

The variation of R 2 as a function of R 1 measured (in newtons) on a fixed ladder is shown in

As discussed in Part I, the force platforms described in the previous section did not permit direct measurement of the parallel force R 4 of the wall on the ladder. Unlike a single beam, which could be positioned with the top at the center of the vertical platform―and therefore directly above the load cell that measures shear―the top of the ladder spanned the width of the platform. Small deviations in position of the ladder resulted in erratic variations in the resulting values for R 4 . Nevertheless, R 4 could be obtained independently from the equation of static equilibrium in the vertical direction: R 4 = P + W − R 1 . Substitution of Equation (3) for R 4 leads to the relation

R 1 + μ R 3 = P + W (10)

which provides an independent test of the assumed frictional force law. In

of the 4 loads P for each of the 6 load locations β . If Equation (3) correctly represents the relation between parallel and normal forces of reaction at the wall, then the resulting plot should be a single straight line of slope 1 upon which all 24 points should closely lie, since μ is expected to be independent of P and β .

Load Location β | Theoretical μ g ( β ) | Empirical μ g ( β ) |
---|---|---|

32/244 | 0.049 | 0.050 ± 0.005 |

62/244 | 0.095 | 0.095 ± 0.005 |

92/244 | 0.143 | 0.150 ± 0.005 |

122/244 | 0.192 | 0.205 ± 0.005 |

152/244 | 0.242 | 0.240 ± 0.005 |

182/244 | 0.294 | 0.300 ± 0.005 |

A more complete and accurate formulation of the law of static friction―i.e. the friction between two dry bodies in the absence of relative motion―takes the form of an inequality. Stated in words [

・ The force of sliding friction between two surfaces relatively at rest is less than or equal to a certain constant times the normal force to the surfaces. (11)

・ The maximum force of friction between the two surfaces, which is the tangential force required to initiate motion, is directly proportional to the normal force. (12)

・ The proportionality constant in statement (12) is the defined coefficient of friction. (13)

Applied to the contact of the ladder at the ground, statements (11)-(13) become

R 2 ≤ F max = μ 0 R 1 (14)

in which F max is the maximum frictional force that the ground can provide, and μ 0 is the coefficient of friction of the ladder with the ground. The equality sign applies at the verge or onset of relative motion of the two surfaces. The physical content of Equation (14) is that if F max does not exceed the parallel force of reaction R 2 (which, by the equations of static equilibrium, is equal to the normal force of reaction R 3 on the ladder by the wall; see

Substitution of Equation (4) for R 2 into Equation (14) leads to the inequality

μ 0 ≥ 1 2 W + β P W ( tan θ + 1 2 μ ) + P ( tan θ + ( 1 − β ) μ ) (15)

which, to insure no sliding, relates the coefficient of friction μ 0 at the ground and the frictional constant μ at the wall. If the wall is frictionless and the ladder carries no load, then Equation (15) reduces to a purely geometric relation

μ 0 ≥ ( 2 tan θ ) − 1 (16)

that is frequently found in elementary mechanics textbooks [

tan θ min = 1 2 W ( 1 − μ g μ ) + P ( β − μ g μ ( 1 − β ) ) μ g ( W + P ) . (17)

The practical utility of relation (15) or (17) is that it determines the angles of inclination that should be safe for a climber of weight P to ascend to a specified height.

In view of the content of Section 2.3, the question posed previously regarding the difference in how the laws of friction were applied at the wall and at the ground is seen to be the wrong question. Rather than asking why an equation of the form of (3) does not apply at the ground, the appropriate enquiry should be why an inequality of the form of (14) does not apply at the wall.

The answer to the right question is that use of Equation (3) at the wall is entirely

consistent with the laws of static friction under the prevailing condition by which equilibrium of the ladder or beam was maintained. In the reported experiments, the ladder did not slide along the ground because friction provided by the ground exceeded the normal reaction R 3 on the ladder by the wall. If the ground were frictionless, the ladder would have slid down the wall because the maximum upward force of friction at the wall was less than the residual downward vertical force W + P − R 1 . Note, however, that the vertical reaction R 4 on the ladder is the maximum force of friction obtainable from the wall―and therefore, according to statement (12), R 4 should be directly proportional to the normal force R 3 at the wall. In other words, Equation (3)―and not an inequality analogous to relation (14)―constitutes a legitimate application of the laws of static friction to the ladder.

It is important to underscore that the theoretical model that was confirmed experimentally in Part I and in the preceding sections of Part II predicts the forces of reaction on a ladder under usual conditions of use. By “usual” is meant that the friction provided by the vertical support (the wall) is insufficient to prevent the ladder from sliding at the ground if the contact between the ladder and the ground were smooth (i.e. nearly frictionless). One can imagine, however, a system―such as a ladder inclined against a very rough stone wall―whereby the maximum force of friction at the wall exceeds the net vertical force downward. Then, if it is still the case that the maximum force of friction at the ground exceeds the parallel reaction R 2 , the laws of static friction become inequalities at both the wall and ground. Such a statically indeterminate system would again be unsolvable without further complementary information because the forces of friction cannot be related to the forces of reaction. Although the coefficients of friction at the wall and ground would determine whether the ladder slid or not, they would not enter, and therefore could not determine, the mathematical expressions for the forces of reaction in static equilibrium.

The author is unaware of any published theory or experiment that examined the static equilibrium of a ladder supported at both base and top by surfaces sufficiently rough that each alone was capable of preventing the ladder from sliding. One approach to analyzing such a system within the framework of boundary constraints introduced in Part I, might be to regard the ladder (or E-B beam) as pinned at both the ground and wall. To recapitulate briefly, the three fundamental kinds of supports commonly encountered in the mechanics of deformable media are roller, pin, and fixed [

The model of a ladder as a single E-B beam pinned at both ends was not included among the 3 models analyzed in Part I because it led to results that could be dismissed immediately as unrepresentative of the statics of a ladder in usual use. These results, however, may apply to the extraordinary condition discussed now, and therefore the analysis is given here. The steps of the analysis are as follows.

1) Transform reactions ( R 1 , R 2 , R 3 , R 4 ) , which are perpendicular or parallel to the ground or wall, into reactions ( A 1 , A 2 , B 1 , B 2 ) , which are axial or transverse with respect to the beam, as shown in

2) Use the equations of static equilibrium to solve for the transverse reactions ( B 1 , B 2 ) . This can be done because there is no unknown internal bending moments at pinned contacts with the ground and wall. (Note: Such moments exist for a fixed contact. See [

3) Write the expression for the axial strain energy. This will contain one of the axial reactions. (In the analysis below, the axial strain energy is expressed in terms of A 2 .)

4) Express the axial reaction in terms of one of the perpendicular or parallel reactions. (In the analysis below, axial reaction A 2 is expressed in terms of perpendicular reaction R 1 .)

5) Use Castigliano’s theorem [

6) From steps 2 and 5 the entire set of reactions ( R 1 , R 2 , R 3 , R 4 ) can be obtained.

From

A 1 = R 1 sin θ + R 2 cos θ B 1 = R 1 cos θ − R 2 sin θ A 2 = − R 3 cos θ + R 4 sin θ B 2 = R 3 sin θ + R 4 cos θ (18)

with inverse relations

R 1 = B 1 cos θ + A 1 sin θ R 2 = − B 1 sin θ + A 2 cos θ R 3 = − B 2 sin θ − A 2 cos θ R 4 = B 2 cos θ + A 2 sin θ (19)

Solution of the equations of static equilibrium for the transverse reactions lead directly to

B 1 = ( 1 2 W + ( 1 − β ) P ) cos θ B 2 = ( 1 2 W + β P ) cos θ (20)

From Equation (20) and the second and fourth relations of Equation (18), one can express R 3 and R 4 in terms of R 1

R 3 = ( R 1 − 1 2 W − ( 1 − β ) P ) cos θ sin θ R 4 = W + P − R 1 (21)

Substitution of Equation (21) into the third relation of Equation (18) leads to the axial reaction A 2 in terms of the perpendicular reaction R 1

A 2 = W ( 1 − 1 2 cos 2 θ ) + P ( 1 − β cos 2 θ ) − R 1 sin θ . (22)

The axial strain energy takes the general form

U a = 1 2 E A ∫ 0 L F a ( x ) 2 d x (23)

where F a ( x ) is the compressive or tensile force throughout the beam. Applying Equation (23) to a ladder modeled as an E-B beam, one finds

U a = L 2 E A [ ∫ 0 β ( A 2 − W ( 1 − ξ ) sin θ − P sin θ ) 2 d ξ + ∫ β 1 ( A 2 − W ( 1 − ξ ) sin θ ) 2 d ξ ] (24)

after making a transformation of variables ξ = x / L . Substitution of Equation (22) into Equation (24) and solution of the equation resulting from Castigliano’s theorem

δ ⊥ ( g ) = ∂ U a ∂ R 1 = 0 , (25)

where δ ⊥ ( g ) is displacement of the contact point at the ground in the direction of force R 1 , leads to

R 1 = 1 2 W + ( 1 − β ) P (26)

from which follows from Equation (21)

R 4 = 1 2 W + β P (27)

and

R 2 = R 3 = 0 . (28)

The axial reactions at the base and top of the ladder

A 1 = ( 1 2 W + ( 1 − β ) P ) sin θ (29)

A 2 = ( 1 2 W + β P ) sin θ (30)

follow from Equation (28) and the first and third relations of Equation (18).

The set of reactions (26)-(28) are independent of the angle of inclination θ , in marked contrast to the forces of reaction on an inclined ladder under usual circumstances. The foregoing solution is likewise obtained if the axial strain energy function (23) were expressed in terms of the parallel reaction R 2 and the application of Castigliano’s theorem took the form

δ ∥ ( g ) = ∂ U a ∂ R 2 = 0 (31)

where δ ∥ ( g ) is displacement of the ground contact point in the direction of R 2 . In fact, the same solution is obtained by applying Castigliano’s theorem directly to the axial strain energy function (24) without making the transformation (22). Although this would have been a simpler way to solve the problem, the approach was not followed here because the corresponding virtual displacement δ axial ( wall ) = ∂ U a / ∂ A 2 or δ axial ( ground ) = ∂ U a / ∂ A 1 would be in a direction not permitted by the ground and wall constraints.

In the paper [

The foregoing proportionality of parallel and normal reactions was shown not to hold at the ground. At the ground, the theory predicted―and the measured reactions confirmed―that the force of friction and the normal force were linearly related by a coefficient that depended sensitively on the position (although not the magnitude) of the load. This apparently asymmetric consequence of friction at the wall and at the ground was resolved by noting that, in general, the law of static friction takes the form of an inequality. The law becomes a linear proportionality only when the frictional force is the maximum obtainable from the immovable surface. This explanation envisions a system, such as examined experimentally in this paper and in Part I, in which a movable object (e.g. ladder) can slide against an immovable object (e.g. wall). It is understood, of course, that the force of friction depends on both surfaces.

If this explanation is correct, then the following conclusions can be drawn from it regarding the role of friction in the experiments reported in Parts I and II:

・ Friction at the ground physically maintains the ladder or beam in static equilibrium by preventing sliding.

・ Friction at the wall (together with the equations of static equilibrium) determines theoretically the four forces of reaction. In other words, the coefficient of friction at the ground does not enter any of the theoretical expressions for the reactions.

・ The reason why friction affected the ladder or beam differently at the ground than at the wall (i.e. “inequality” vs. “proportionality”) is due to the fact that the magnitude of friction, relative to other acting forces, was different at those two supports. At the ground, the maximum obtainable friction was greater than the parallel reaction R 2 . At the wall, the maximum obtainable friction (identical to the parallel reaction R 4 ) was less than the net downward force.

・ In light of the preceding comments, the frictional constant μ in Equation (3) is the canonically defined coefficient of friction at the wall―and therefore should yield (within experimental error) the same numerical value independent of load and load placement whether deduced from the forces of reaction on the ladder or measured by any of the standard methods for determining coefficients of static friction [

The forces of reaction derived and tested in Parts I and II pertained to a physical system in which the force of friction at the wall was insufficient by itself to maintain the ladder in static equilibrium, i.e. to keep the ladder from sliding if the ground surface was smooth. Under the uncommon circumstance where a ladder is supported by two rough surfaces (ground and wall) at which neither the top nor base of the ladder is on the verge of sliding, the effect of friction at each surface must be expressed by an inequality. This statically indeterminate problem would then require additional complementary information to be solvable. In the model analyzed in Section 2.4 of a single-beam ladder with both ends pinned, this complementary information made use of the axial strain energy function of the beam, although the resulting forces of reaction were found not to depend on the elastic constant E.

The author is unaware of any experimental test of such a system. Whether the model of an inclined Euler-Bernoulli beam pinned at both ends satisfactorily describes a ladder in static equilibrium under the hypothetical condition portrayed above can be decided only by an experimental test. Such a test is outside the scope of this paper.

E. Donovan, W. Emery, J. Hallquist, W. Scully, L. Sewanan, and W. Strange each helped take data at some point during the initial phase of this project. The author also thanks Trinity College for partial support through the research fund associated with the George A. Jarvis Chair of Physics.

The author declares no conflicts of interest regarding the publication of this paper.

Silverman, M.P. (2018) The Role of Friction in the Static Equilibrium of a Fixed Ladder: Theoretical Analysis and Experimental Test. World Journal of Mechanics, 8, 445-463. https://doi.org/10.4236/wjm.2018.812032