Friction and Hydrostatic Forces - Study Notes

FRICTION

• Objective:

  • a) Understand the characteristics of dry friction.

  • b) Draw a free body diagram (FBD) including friction.

  • c) Solve problems involving friction.

• Key definitions:

  • Friction is a force of resistance acting on a body which prevents or retards slipping relative to a second body.

  • Friction acts tangent (parallel) to the contacting surface and opposite to the relative motion or tendency for motion.

  • For equilibrium in the shown setup: $F = P,<br>N = W,<br>Wx = Ph.$ (illustrative balance relations in a typical incline/slider problem)

• Direction of friction:

  • The friction force always acts to oppose motion or impending motion; it is directed along the contact surface and opposite to the tendency to slip.

• Practical questions:

  • In braking designs, how do we determine the magnitude and direction of the friction force given an applied force on brake pads?

• Major concepts: static vs kinetic friction

  • Maximum static friction: Fs = s N where s = \mus is the coefficient of static friction.

  • If the block begins to move, kinetic friction applies: $F<em>k = \mu</em>k N$ where \muk < \mus.

  • Frictional force varies with the applied load; the maximum static friction occurs just before impending motion.

• Determining static friction experimentally:

  • Tilted plane method: place block on an inclined plane, gradually tilt until just about to slip.

  • At impending motion: $\mu<em>s = \frac{W \sin \theta</em>s}{W \cos \theta<em>s} = \tan \theta</em>s$

  • Equilibrium relations at impending motion:

  • $\sum F<em>y = N - W\cos\theta</em>s = 0$

  • $\sum F<em>x = \mu</em>s N - W\sin\theta_s = 0$

• Impending tipping vs slipping:

  • For a given weight W and height h, determine whether the block will slide or tip first.

  • There are four unknowns (F, N, x, P) and only three equations of equilibrium (EoE), so an assumption is needed to obtain a fourth equation.

  • Two common approaches:

  • Impending tipping assumption:

    • Known: $F = \mu_s N$

    • Solve for $x, P, N$ and check: $0 \le x \le \dfrac{b}{2}$

  • Impending tipping alternative:

    • Known: $x = \dfrac{b}{2}$

    • Solve for $P, N, F$ and check: $F \le \mu_s N$

• Example: Ladder problem (friction at floor, smooth wall)

  • Given: a uniform ladder weighs 100 N; vertical wall is smooth (no friction); floor is rough with $\mu_s = 0.8$.

  • Question: Find the minimum force $P$ needed to move (tip or slide) the ladder.

  • Plan:

  • a) Draw an FBD of the ladder.

  • b) Determine the unknowns.

  • c) Make any necessary friction assumptions.

  • d) Apply equations of equilibrium and friction equations to solve for unknowns.

  • e) Check the assumptions, if required.

  • Geometry from the figure (as given):

  • Ladder length components include a horizontal span of 2 m and vertical height components 1.5 m each (as shown in the diagram).

  • Known/unknowns (as per plan): $W = 100\ ext{N},\ \mu_s = 0.8,$ unknowns include the external push $P$, normal reaction $N$, and friction force at the floor $F$ at the contact point with the floor.

  • FBD notes: include friction at the floor opposing motion, and the normal reaction at the floor and at the wall (the wall is smooth so only normal reaction from wall).

• Example: Drum group problem (impending motion)

  • Given: Drum weight = 500 N, $\mu_s = 0.5$, geometry $a = 0.75\,\text{m}, b = 1\,\text{m}$.

  • Task: Find the smallest magnitude of $P$ that will cause impending motion (tipping or slipping).

  • Plan:

  • a) Draw a FBD of the drum.

  • b) Determine the unknowns.

  • c) Make friction assumptions as necessary.

  • d) Apply EoE (and friction equations as appropriate) to solve for unknowns.

  • e) Check assumptions.

  • FBD notes: labels show internal points X3, X4, X5 and a lever arm for the push P; the diagram indicates the drum with a base width b, height parameters, and the reactions N and friction F at contact.

• FBD guidelines (summary):

  • Always draw friction opposing motion or impending motion.

  • Do not automatically set $F = \mu_s N$ unless impending motion is explicitly given.

  • Use equilibrium equations plus friction relations to solve for unknowns.

HYDROSTATIC FORCES

• Fluid statics vs fluid dynamics

  • Fluid statics deals with fluids at rest; fluid dynamics studies fluids in motion.

  • A fluid at rest has no shear stress; any force is due to normal stresses (pressure). This is the hydrostatic condition.

• Importance and applications of hydrostatics

  • Atmosphere and oceans can be approximated as at rest in many problems.

  • Applications include determining pressures at different atmospheric levels, forces on submerged objects (ships, subs, dams, hydraulic structures), and pressure-measuring devices.

  • Hydrostatic concepts can be generalized to other liquids, but the discussion here focuses on water.

• Free surface and pressures

  • The surface of water in contact with air is the FREE SURFACE.

  • Pressure at the free surface is atmospheric; absolute pressure is the sum of gauge pressure and atmospheric pressure:
    $p<em>{abs} = p</em>{g} + p_{atm}.$

  • For a pool/t tank: pressure increases with depth and is given by
    $p = \rho g h,$ where

  • $\rho$ is the density of the liquid (for water, typically ~1000 kg/m^3),

  • $g$ is the acceleration due to gravity (approx. 9.81 m/s^2),

  • $h$ is the depth below the free surface.

  • Pressure acts perpendicular to submerged surfaces (normal to the surface).

• Notation and constants

  • Water unit weight $\gamma = \rho g \approx 9.81\ \text{kN/m}^3$ (often approximated as 9.8 kN/m^3 in problems).

  • Example substitution in problem statements may use 9.8 for simplicity.

• Example 1: Hydrostatic force on a square plate at the bottom of a tank

  • Given: Tank with water depth 8 m, plate size 4 m × 4 m, plate at bottom.

  • Pressure on plate (uniform, since depth is constant across the plate):
    $p = \rho g h = 9.8\,\frac{\text{kN}}{\text{m}^3} \times 8\,\text{m} = 78.48\,\frac{\text{kN}}{\text{m}^2}.$

  • Force on plate:
    $F = p A = 78.48\,\frac{\text{kN}}{\text{m}^2} \times (4\,\text{m} \times 4\,\text{m}) = 1255.68\,\text{kN}.$ (≈ 1255.7 kN)

  • Result: The hydrostatic force on the plate is equivalent to a single resultant normal force acting at the centroid of the plate, perpendicular to the plate.

• Example 2: Hydrostatic force on a vertical wall with nonuniform pressure distribution

  • Given: Tank width 1 m; water depth 4 m; wall ABCD in contact with water.

  • Pressure varies with depth: at depth h, $p = \rho g h = 9.8 \times h$ (kN/m^2).

  • Maximum pressure at bottom:
    $p<em>{max} = \rho g h</em>{max} = 9.8 \times 4 = 39.2\ \text{kN/m}^2.$

  • For a wall of width 1 m, the line load (per meter width) is a triangular distribution with base height 4 m and top value 0 at the surface.

  • The load is nonuniform, and the resultant force is the area under the triangle:
    $R = \frac{1}{2} p_{max} h = \frac{1}{2} (39.2) (4) = 78.4\ \text{kN} \text{ per meter width}.$

  • Location of the resultant: the centroid of the triangle, located one-third of the height above the base (from the bottom):
    $y = \frac{1}{3} h = \frac{1}{3} (4) = 1.33\ \text{m}.$

  • Graphical representation shows a nonuniform distribution on the vertical wall, forming a triangle with base along the depth direction.

• Key takeaways about hydrostatics:

  • Pressure increases linearly with depth: $p = \rho g h$.

  • For uniform depth across a submerged plate, the pressure is uniform across the plate surface.

  • For vertical walls with depth varying over the height, the pressure distribution is triangular, leading to a resultant force located at a depth of h/3 from the surface (or equivalently h - h/3 from the bottom, depending on reference).

  • The resultant hydrostatic force on a submerged surface can be found by integrating the pressure distribution or using geometric area formulas for simple distributions (rectangular, triangular, trapezoidal).

• Nonuniform vs uniform pressure distribution (visual note):

  • Nonuniform pressure distribution: pressure varies with depth, resulting in a triangular (or more complex) loading pattern on submerged surfaces.

  • Uniform pressure distribution: pressure is the same across the surface, leading to a single resultant equal to the pressure times the area.

• Summary of formulas (for quick reference):

  • Absolute pressure: $p<em>{abs} = p</em>{g} + p_{atm}$

  • Gauge pressure in a fluid at rest: $p = \rho g h$

  • Resultant force on a vertical wall with height h and width b for a triangular distribution: $R = \frac{1}{2} p_{max} h = \frac{1}{2} (\rho g h) h \, b = \frac{1}{2} \rho g h^2 \; b$

  • Position of resultant for triangular distribution: at depth $y = \frac{h}{3}$ from the free surface (or equivalently 2h/3 from the surface depending on reference).

• Notes on units:

  • Pressure units: $\text{kN/m}^2$ (or kPa).

  • Force units: $\text{kN}$.

  • Depths in meters, width/area in square meters.

• These notes provide a compact reference to the key ideas, equations, and example calculations for friction and hydrostatic forces, including practical problem-solving steps and common pitfalls (e.g., assuming always that F = \mu_s N without checking the impending motion condition).