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The angle of repose characterizes the free-surface slope of a resting granular material, measured relative to horizontal under a specified preparation protocol. A slowly poured dry sand pile often forms an approximately conical surface, with slope $\tan {\theta }_r$. It is a bulk consequence of contact friction, particle shape, packing and any cohesion, rather than a universal constant of one chemical substance.

A simple comparison is to pour a pile through a funnel, measure its height and base radius, and infer the slope from their ratio. A second experiment slowly tilts a layer until it avalanches. The onset angle and the angle after flow stops can differ, so reports should state which quantity was measured. For a single sliding block, static balance suggests $\tan \theta =\mu$, but identifying a pile's angle of repose with a microscopic coefficient of friction ignores rearrangement and collective contact geometry.

The contact slip displacement is signed relative tangential displacement at a specified contact. For small motions in a fixed local frame, let $\mathbf{t}$ be a tangential unit vector and let ${\mathbf{u}}_1,{\mathbf{u}}_2$ describe the two contacting surfaces. ThenThis scalar product extracts relative motion along the contact. The corresponding relative normal displacement is a separate quantity controlling opening or closure. This local distinction matters for inter-fibre sliding: an entire bundle can move without adjacent material fibres slipping against one another.

As an original local example, take $\mathbf{t}=(1,0)$, ${\mathbf{u}}_1=(0.30,0.02)\mathrm{mm}$ and ${\mathbf{u}}_2=(0.10,0.02)\mathrm{mm}$. The contact slip displacement is $0.20\mathrm{mm}$ and the relative normal displacement is 0. Adding the same translation to both surfaces changes neither result. Reversing the orientation of $\mathbf{t}$ reverses the sign of ${\delta }_s$ while preserving its magnitude. If the contact frame rotates substantially or different material points enter contact, relative tangential velocity must be evaluated in the current contact frame and integrated consistently; projecting one final global displacement difference onto a final tangent vector is not a general finite-motion formula.

Under the ideal dry-friction model, static friction can transmit tangential force while the local slip rate is 0. Sliding introduces kinetic friction; with constant normal force $N$, its magnitude is ${\mu }_{k}N$, where ${\mu }_k$ is the coefficient of friction.^[ref] The accumulated contact slip and the frictional dissipation from contact slip are therefore different from net signed displacement:The second expression follows from integrating the opposing frictional force over each increment of relative sliding; it assumes constant $N$ and ${\mu }_k$. For a prescribed out-and-back slip from 0 to $1\mathrm{mm}$ and back to 0, the final ${\delta }_s$ is 0 but ${\ell }_s=2\mathrm{mm}$. With stipulated $N=0.20\mathrm{N}$ and ${\mu }_k=0.25$, the dissipated energy is $100\mu \mathrm{J}$. Zero net slip does not imply zero frictional dissipation.

A simple measurement marks both surfaces close to a maintained contact, images their relative positions while tangential loading is varied, and resolves motion along the local contact direction. Measurement calibration converts image distances to lengths. Marks far from the contact can also move because the strands stretch or the fixtures deform; a grip-travel readout alone does not isolate contact slip displacement. Record both the signed motion and accumulated contact slip when comparing repeated bending cycles of an unbonded material-fibre bundle.

Coulomb friction is an approximate contact model in which tangential friction is limited by a coefficient of friction times compressive normal load. For a dry contact with normal force $N\ge 0$, the usual static and sliding versions arewhere $\mathbf{w}$ is contact slip. Static friction adjusts within its bound; it is not always equal to the limiting value. Kinetic friction opposes the relative sliding at the contact, which need not be opposite to the body's centre-of-mass velocity.^[ref]

For a horizontal contact with $N=100\mathrm{N}$ and stipulated ${\mu }_s=0.4$, an applied tangential force of $20\mathrm{N}$ can be balanced while sticking; the limiting value is $40\mathrm{N}$. In geological applications, a cohesive shear threshold is often added to the load-dependent term, with effective normal stress reduced by pore pressure. Real contact behavior can also depend on sliding history, speed and adhesion, so these constants describe a stated regime.

An end effector is the device at a robot arm's working end that directly performs its task: for example, a gripper, welding tool or suction cup. The arm places and orients it, while the tool's geometry and contact behavior determine how that motion affects the object. A tool-tip frame can differ from the mounting-flange frame, so its offset must be included in forward kinematics.^[ref]

For example, suppose a two-finger gripper holds an object vertically by dry friction. If each finger applies normal force $N$ and the coefficient of friction is $\mu$, the ideal upward friction capacity is $2\mu N$. With $\mu =0.5$ and $N=10\mathrm{N}$ per finger, that capacity is $10\mathrm{N}$, just above the weight of a one-kilogram object near Earth's surface. A real test varies gripping force and load while observing slip; contact material, deformation and acceleration can invalidate the simple estimate. Positioning accuracy alone therefore does not establish grasp reliability.

Gear sliding-friction loss is the rate at which relative tooth-flank sliding dissipates mechanical energy. In a simple Coulomb friction model with normal contact load $F_n$, coefficient of friction $\mu$ and gear sliding velocity $v_s$, the nonnegative dissipated power isThis is a local contact model, rather than the entire gearbox loss. Manufacturer guidance distinguishes gear-meshing losses from losses in rotating supports and warns that overall gear efficiency depends on load and speed.^[ref]

To derive the sign, let $v_s=v_1-v_2$ along the common tangent. The friction force on surface1 is $f_1=-\mu F_n\mathrm{sgn}(v_s)$ and that on surface2 is $-f_1$. Their combined mechanical power is $f_1{v}_1-f_1{v}_2=-\mu F_n|v_s|$. The corresponding positive quantity is the rate of mechanical energy loss. At zero sliding the ideal Coulomb sliding-loss term vanishes, even though another resistance mechanism can remain.

For a fixed-axis external spur gear pair with $|{\omega }_1|+|{\omega }_2|=75rad/s$, $|s|=4\mathrm{mm}$ gives $|v_s|=0.30m/s$. Hypothetical $\mu =0.05$ and $F_n=100\mathrm{N}$ give $P_f=1.5\mathrm{W}$. Holding these load and friction assumptions fixed while doubling both shaft speeds doubles the instantaneous loss at that geometric point.

Gear friction work per engagement integrates this varying loss through the active gear path of contact. If one pair carries a constant $F_n$, the contact coordinate runs from $-L_a$ to $L_r$ at constant speed $\dot{s}=u_b>0$, givingFor $L_a=5.0586\mathrm{mm}$, $L_r=4.5960\mathrm{mm}$ and $u_b=0.93969m/s$, the preceding loads and speeds give about $9.32\mathrm{mJ}$. Real tooth pairs share varying loads; an overall loss calculation must sum the applicable contacts and integrate their actual loads, sliding and friction model.

Gear tooth sliding is relative tangential motion of mating tooth surfaces at their contact point. Ideal contact prevents relative normal penetration but does not generally require the tangential surface velocities to agree. For a fixed-axis external involute pair, tangential slip vanishes at the gear pitch point and changes as the contact moves along the gear path of contact.^[ref]

Let $P$ be the pitch point, let $\mathbf{n}$ be a unit vector along the gear line of action, and write a contact location as $Q=P+s\mathbf{n}$. With a common positive axis $\hat{\mathbf{z}}$ and signed angular velocities ${\omega }_1,{\omega }_2$, rigid-body velocity differences giveThe velocities at $P$ cancel, leaving this term proportional to distance along the contact line. Since ${\omega }_2$ has opposite sign in an external pair, the sliding-speed magnitude is $(|{\omega }_1|+|{\omega }_2|)|s|$.

For an original example, use ${\omega }_1=50rad/s$ and ${\omega }_2=-25rad/s$. At $s=4\mathrm{mm}$, gear sliding velocity along the common tooth tangent is $0.30m/s$; at $s=-4\mathrm{mm}$ it has the opposite sign. At $s=0$ it is zero. Equality of operating pitch-circle speeds must therefore not be used to conclude that the tooth flanks roll without slipping throughout engagement.

In a simplified Coulomb friction model with coefficient of friction $\mu$ and normal load $F_n$, the dissipated gear sliding-friction loss is $P_f=\mu F_n|v_s|$. Hypothetical values $\mu =0.05$, $F_n=100\mathrm{N}$ and $|v_s|=0.30m/s$ give $1.5\mathrm{W}$. This is a local illustrative loss model, not a lubricant specification. Zero ideal sliding at the pitch point also does not eliminate every source of gearbox loss, such as support friction or deformation-related dissipation. Real load and friction can vary through the engagement, so total loss requires integrating the applicable contact model.

friction force microscopy is suitable. Contact scanning twists the AFM cantilever; the lateral photodetector difference measures friction and can distinguish the supported graphene and glass even if their heights are close. Acquire both trace and retrace and use half their lateral-signal difference to suppress slope-induced cross-talk. Keep the normal load low enough to avoid dragging or damaging the flake, control humidity and contamination, and calibrate torsional sensitivity if a quantitative coefficient of friction is required. An electrical connection is unnecessary.

Write the sphere's mass as $M$ and use $I=2M{a}^2/5$ from b. Let $K_n$ and $K_t$ be the upward and rightward contact impulses. The initial vertical velocity is $-u_1$ and the outgoing vertical velocity is $u_2$. Perfect normal elasticity givesThe tangential and angular impulse equations areFor the ideal elastic rough-sphere bounce, the tangential translational plus rotational kinetic energy is restored on separation. Substituting these two equations into that requirement givesThe interacting rough-contact branch, rather than the smooth branch $K_t=0$, therefore hasConsequently, with all tangential quantities signed in the convention of c,In particular, $v_2+a{\Omega }_2=-(v_1+a{\Omega }_1)$: the tangential contact-point velocity reverses. The outgoing $v_2$ may be negative; the sphere rebounds backward when $a{\Omega }_1>3v_1/4$. Substitution also checks both total kinetic energy and angular momentum about the contact point, $I\Omega -Mav$.

This result uses the conventional perfectly elastic, fully rough contact idealization. The word rough alone, with an unspecified finite coefficient of friction, would not determine the outgoing tangential state. For example, imposing sticking at separation would instead give $K_t=-2M(v_1+a{\Omega }_1)/7$ and dissipate $M(v_1+a{\Omega }_1{)}^2/7$, contradicting total elasticity unless $w_1=0$. The displayed elastic branch requires enough tangential impulse; a Coulomb bound $|K_t|\le \mu K_n$ would at least require $\mu \ge 2|v_1+a{\Omega }_1|/(7u_1)$.

Use a rock strength envelope: at each depth plot the smaller differential stress needed for frictional failure and for sustained ductile material creep at a specified strain rate. Shallow rock generally strengthens with depth because effective confining pressure increases; sufficiently deep hot rock weakens because thermally activated material creep becomes efficient.

At shallow depth, Byerlee's law or a Mohr-Coulomb failure criterion gives a relation of the formwhere $C$ is cohesion, $\mu$ a coefficient of friction, ${\sigma }_n$ the compressive normal stress and $P_f$ the pore pressure. Increasing overburden generally increases effective stress, hence the frictional strength. Conversion from fault shear stress to differential stress depends on the stress regime and fault orientation; a sketch should not label a universal numerical slope. High pore pressure or weak fault gouge can lower the frictional strength substantially.

A simplified dislocation creep relation isHere $Q$ is activation energy, $V^{\ast }$ activation volume, $R$ the molar gas constant and $T$ absolute temperature; $A$ and $n$ depend on composition and deformation mechanism. At fixed strain rate, increasing temperature usually decreases the required stress strongly. The competing effects of pressure, water weakening of rocks, grain size and composition also matter. The intersection with frictional failure gives a brittle-ductile transition, not a change from solid to liquid.

For the Indian continental interior, a useful first sketch has an upper-crustal strength peak, a weaker lower continental crust, and renewed strength in lithospheric mantle beneath the Moho. Quartz-bearing rocks can undergo ductile material creep at conditions where colder, drier olivine-rich mantle retains appreciable strength. This creates a possible two-peak, or “jelly sandwich”, rock strength envelope. With further depth the mantle also weakens. A cold craton can retain strong lithospheric mantle to much greater depths than a tectonically heated continental region.

For the central Pacific, show thin mafic oceanic crust above a relatively thick layer of strong lithospheric mantle. There need not be a conspicuous weak lower-crustal layer. Much of the integrated strength can reside below the Moho, with a broad strength peak followed by ductile weakening towards the asthenosphere. Cooling and age control the thickness: young hot oceanic lithosphere would give a much shallower and weaker profile than an old central-Pacific plate.

The plotted Moho depths of about 38 km and 7 km, and all plotted stresses, are illustrative assumptions. “India” alone does not specify a unique geothermal gradient, water content or strain rate, and the Pacific includes plates of different ages. The robust comparison is the possible crust–mantle layering in a continent versus the thin oceanic crust and mantle-dominated strength of an old oceanic plate. The Moho is a compositional boundary; neither it nor the deepest small plotted strength is automatically the seismic lithosphere-asthenosphere boundary.

Left: ordinary frictional strengthening with depth followed by thermally activated material creep. Right: a very weak shallow faulted layer above a narrow strong region near 10 km, followed by essentially the same weakening material creep branch. These are the shapes actually plotted in Figure 2. The right curve is not a strong upper crust over a uniformly weak lower crust; neither diagram displays a second mantle-strength peak.

For the left curve, increasing overburden increases the effective normal load on faults. A schematic Coulomb friction law iswhere $C$ is cohesion, $\mu$ the coefficient of friction, ${\sigma }_n$ the compressive normal stress and $P_f$ the pore pressure. With ordinary rock friction and approximately hydrostatic pressure, ${\sigma }_n-P_f$ increases roughly linearly with depth. The stress needed for frictional sliding therefore increases, as in the straight shallow branch. Byerlee's law summarizes the relatively high friction found in many laboratory rocks.^[ref]

At higher temperature, dislocation creep or another ductile mechanism accommodates the imposed deformation at lower stress. For example,Here $\dot{ϵ}$ is strain rate, $Q$ an activation energy, $R$ the molar gas constant and $T$ absolute temperature. At fixed strain rate, increasing $T$ lowers the required stress. The operative envelope is approximately $\min ({\sigma }_f,{\sigma }_d)$, expressed in a consistent stress convention; the peak marks the brittle-ductile transition. This accounts for the left curve's peak near 100 MPa and its decline toward small strength by about 20 km.

For the right curve, normal high-friction sliding with depth-independent fluid-pressure ratio cannot explain the nearly vertical weak shallow branch. Plausible weakening includes high pore pressure tracking the overburden, low-friction fault gouge, or a connected network of pre-existing weak faults. Roughly constant small ${\sigma }_n-P_f$ yields roughly constant small frictional strength. The abrupt increase near the base of this layer requires a change of conditions: for example, the weak fluid-rich fault network terminates or becomes discontinuous, leaving stronger intact or less altered rock below. A change in mineralogy or deformation localization can also give a contrast. Simply saying “there are fluids” does not explain the return to high strength. Below the strong region, thermal material creep again lowers the strength.

The figure alone does not identify the weakening mechanism or a particular lithological boundary. In particular, a 10 km strength maximum is not evidence for a 10 km deep Mohorovičić discontinuity. The strong-fault/weak-fault evidence and the distinction between fault strength and surrounding crustal strength are discussed in Stress orientations at intermediate angles to the San Andreas Fault, California by Jeanne L. Hardebeck and Andrew J. Michael.

Support for the left curve comes from high laboratory friction and substantial in situ differential stress; support for the right curve comes from low inferred fault traction, weak fault materials or high pore pressure. No single observation directly measures the whole plotted envelope.

For the left, stronger shallow-crust model:

For the right, weak shallow-layer model:

To discriminate, measure the strength-controlling quantities through depth and across both faults and their surroundings. Obtain borehole stress, pore pressure and temperature profiles where feasible; use borehole breakouts and hydraulic tests to constrain effective stress. Recover fault and host rocks and test their friction and material creep at relevant fluid conditions. The right curve specifically predicts an end to the shallow weakening near its sharp rise: investigate changes in permeability of a porous medium, alteration and fault gouge with depth, rather than testing only a surface sample.

Combine those measurements with heat flux transects, magnetotelluric sounding for conductive fluids, accurate earthquake depths and focal mechanisms, and GNSS or InSAR observations of deformation. The latter can help constrain the depth distribution of locking and postseismic relaxation, but neither locking nor seismicity alone gives absolute strength. Earthquake stress drop is a change in stress, not the pre-event stress. Low heat flux can also reflect heat transport or transient weakening; high electrical conductivity can have causes other than free water. A joint explanation of all these data is needed to choose between the models locally.

A numerical answer needs an assumption about fault orientation. For a cohesionless fault optimally oriented for Mohr-Coulomb failure criterion, with vertical ${\sigma }_1$, write $\phi$ for the internal friction angle and $\delta$ for the fault dip. The angle between the plane and ${\sigma }_1$ is $90^{\circ }-\delta$, so the optimal normal-fault dip relation givesHence ${\mu }_f=\tan 40^{\circ }=0.839\simeq 0.84$ under that model.

For the Mohr circle, plot effective compression ${\sigma }_n^{\prime }$ horizontally and shear stress $\tau$ vertically. Draw a circle with diameter $[{\sigma }_3^{\prime },{\sigma }_1^{\prime }]$ tangent to $\tau ={\mu }_f{\sigma }_n^{\prime }$. Its centre and radius obey $R/C=\sin \phi$. One illustrative normalization is ${\sigma }_3^{\prime }=1$, ${\sigma }_1^{\prime }=4.59891$, with tangency at the resolved stresses on the $65^{\circ }$ plane. Pore fluid pressure shifts the total-stress circle horizontally; the friction law uses effective stress.

The dip does not determine the actual sliding coefficient of friction without the optimal-orientation assumption. A reactivated inherited plane may be nonoptimal, cohesion may matter, and the stress ratio and pore fluid pressure are not measured. For an arbitrary compression-positive stress state,at cohesionless sliding. Thus the requested single value is a conditional inference, not a friction measurement.

Use the same conditional optimal normal-fault dip relation as in question 1, with $\delta =55^{\circ }$:The second panel of the original Mohr circle figure has ${\sigma }_3^{\prime }=1$ and ${\sigma }_1^{\prime }=2.03961$; its tangent from the origin has slope $0.364$. Resolving the stresses onto a $55^{\circ }$ plane places its stress point at that tangency. The diagram assumes negligible cohesion, a vertical maximum principal stress and an optimally oriented new or equivalently oriented sliding plane. Without those assumptions, a dip and a statement that sliding began do not uniquely determine the coefficient of friction.

Assume an initially unfractured homogeneous isotropic material described by a linear Mohr-Coulomb failure criterion, with the given coefficient of friction ${\mu }_f=0.5$ representing the relevant internal friction angle. Use spatially uniform pore fluid pressure, effective stresses ${\sigma }_i^{\prime }={\sigma }_i-P_f$, and compression-positive signs. ThenThe Mohr circle in the ${\sigma }_1^{\prime },{\sigma }_3^{\prime }$ plane has centre $C=({\sigma }_1^{\prime }+{\sigma }_3^{\prime })/2$ and radius $R=({\sigma }_1^{\prime }-{\sigma }_3^{\prime })/2$. At first failure it is tangent to the failure envelope. The radius to the positive-shear tangent point is perpendicular to the line of slope $\tan \phi$, hence makes angle $90^{\circ }+\phi$ with the positive normal-stress axis.

If $\theta$ is the angle between the fault normal and ${\sigma }_1$, the stress transformation isThus $2\theta =90^{\circ }+\phi$, $\theta =58.2825^{\circ }$, and the angle of the fault itself to horizontal ${\sigma }_1$ isThe two conjugate thrust faults therefore strike approximately east–west and dip about $32^{\circ }$ north and $32^{\circ }$ south. This is the fault dip, not the $58.3^{\circ }$ fault-normal angle. Both sediments and basement have the same optimum orientation under these assumptions because their stipulated friction slopes are equal; their cohesion and absolute failure stresses could still differ.

The first two panels in solution give the labeled construction. Constant cohesion shifts the straight failure envelope without changing its slope, and uniform pore fluid pressure translates the Mohr circle in normal-stress coordinates; neither changes this tangency angle. In the illustrative cohesionless case, $R/C=\sin \phi$, giving ${\sigma }_1^{\prime }/{\sigma }_3^{\prime }=(1+\sin \phi )/(1-\sin \phi )\simeq 2.618$. Pre-existing weaknesses, a nonuniform stress field, anisotropic strength, or a nonlinear failure envelope invalidate the simple optimum-orientation prediction.^[ref]

The $31.7^{\circ }$ result predicts initial failure in a uniform material under the assumed principal-stress orientations. The actual Figure 3 is a present-day NW–SE cross section through a thick, heterogeneous Alpine stack, extending to about 50 km depth. It contains near-horizontal basal stretches, curved ramps, tightly bent contacts, and steep structures in the internal zones around Viso and the Dora Maira complex. It does not show a single undeformed conjugate pair newly formed in horizontal uniform rock.

Several processes explain the departures:

Finally, a section shows apparent dip unless it is perpendicular to fault strike. For an ideal plane, $\tan {\delta }_{\mathrm{app}}=\tan {\delta }_{\mathrm{true}}\sin \alpha$, where $\alpha$ is the angle between the section direction and fault strike. Obliquity can make a plane look shallower, but cannot on its own explain an apparent dip steeper than its true dip. Do not infer that each black curve is a currently active brittle earthquake fault.