Examples on Dry Friction - Engineering Mechanics - Civil Engineering (CE)

Dry friction occurs when two solid surfaces are in contact and there is a tendency for relative motion but the surfaces are not separated by a lubricant film. This chapter gives clear definitions, fundamental laws, useful formulae, and worked examples suitable for students of engineering and competitive examinations. Emphasis is on static (limiting) friction and kinetic (sliding) friction, their role in equilibrium, and typical problems encountered in structures and machines.

Basic concepts

• Dry friction is the tangential resistance to relative motion between two contacting dry solids.
• Normal reaction (N) is the contact force perpendicular to the surfaces at the interface.
• Frictional force (f) is the tangential force at the interface that opposes relative motion or impending motion.
• Static friction acts when there is no relative motion. It can take any value up to a maximum called the limiting (static) friction.
• Kinetic (dynamic or sliding) friction acts when surfaces slide relative to each other; it is usually less than limiting static friction.

Laws of dry friction

• Frictional force acts tangentially to the common surface and opposite to the direction of impending or actual motion.
• The frictional force is proportional to the normal reaction at the contact: f ∝ N.
• The frictional force is (approximately) independent of the apparent area of contact and of small changes in relative speed for dry contact.
• For static friction, f ≤ μs N, where μs is the coefficient of static friction; the equality f = μs N holds for impending motion and is called the limiting friction.
• For kinetic friction, f = μk N, where μk is the coefficient of kinetic friction, and typically μk < μs.

Mathematical statements

• Limiting (static) friction: fmax = μs N.
• For impending motion: f = μs N.
• Kinetic friction: f = μk N.
• Angle of repose (φ) is the maximum angle of an inclined plane at which a body placed on it will remain at rest without sliding.

Angle of repose and limiting friction

The angle of repose φ is related to the coefficient of limiting friction by the relation:

tan φ = μs

Derivation:

Consider a block just on the verge of sliding on an incline of angle φ. The component of weight down the plane is W sin φ and the normal reaction is W cos φ.

At impending motion downwards, limiting friction f = μs N.

Therefore, W sin φ = μs W cos φ.

Hence, tan φ = μs.

Direction and point of application of friction

• Friction acts at the contact surface and is distributed; when modelling as a concentrated force, it is applied at the point or line where resultant normal reaction acts.
• Friction always opposes the direction of relative motion or impending motion of the body.

Equilibrium conditions involving friction

To determine whether a body will remain in equilibrium on a surface with friction, apply resolution of forces and moments together with the inequality for static friction f ≤ μs N. If equality is reached, the body is at the limit of impending motion.

Worked examples on dry friction

Example 1 - Block on a horizontal plane

Problem. A block of weight W rests on a rough horizontal plane. A horizontal force P is applied to the block. The coefficient of static friction between block and plane is μs. Find the minimum value of P required to just start the block moving.

Sol.

For impending motion, frictional resistance f reaches limiting value f = μs N.

The normal reaction N equals the weight W (no vertical component of P in this problem):

N = W.

Horizontal equilibrium at impending motion gives:

P = f = μs N.

Substitute N = W:

P = μs W.

Therefore the minimum horizontal force to start motion is P = μs W.

Example 2 - Block on an inclined plane (angle of repose)

Problem. A block of weight W rests on an incline of angle θ. The coefficient of static friction is μs. Determine whether the block will slide, remain at rest, or be at incipient motion. Find the critical angle (angle of repose) at which sliding begins.

Sol.

Resolve weight into components along and perpendicular to the plane.

Component down the plane = W sin θ.

Normal reaction N = W cos θ.

Maximum static friction available = μs N = μs W cos θ.

If W sin θ < μs W cos θ, then W sin θ is resisted and the body remains at rest.

If W sin θ = μs W cos θ, the body is at impending motion.

If W sin θ > μs W cos θ, the body will slide downwards.

Critical condition gives:

tan θ = μs.

Thus θcrit = arctan(μs), the angle of repose.

Example 3 - A block pulled at an angle (pulling up or pushing down)

Problem. A block of weight W is pulled by a force P applied at an angle α above the horizontal. The block rests on a rough horizontal surface with coefficient of static friction μs. Determine the minimum magnitude of P to start the block moving, and discuss how α affects P.

Sol.

Resolve P into horizontal and vertical components.

P cos α acts horizontally and tends to move the block.

P sin α acts vertically upward and reduces the normal reaction.

Normal reaction N = W - P sin α.

Limiting friction f = μs N = μs (W - P sin α).

At impending motion, horizontal equilibrium gives:

P cos α = μs (W - P sin α).

Solve for P:

P cos α = μs W - μs P sin α.

P (cos α + μs sin α) = μs W.

P = μs W / (cos α + μs sin α).

As α increases (pulling more upwards), the denominator increases and required P decreases until a limit where P sin α = W (lifting off), beyond which the normal becomes zero and problem changes.

Example 4 - Ladder leaning against a rough vertical wall

Problem. A uniform ladder of length L and weight W rests with its lower end on a rough horizontal floor and its upper end against a rough vertical wall. The coefficient of static friction at both contacts is μs. Find the minimum angle θ with the horizontal for which the ladder will not slip.

Sol.

Take the ladder in static equilibrium. Let reaction at the floor be Rf with normal component Nf and friction Ff; at the wall reaction be Rw with normal component Nw and friction Fw.

Normal reaction at floor Nf = reaction perpendicular to floor = Rf (vertical component balances vertical loads together with wall friction).

For simplicity, consider forces and moments about the lower end to eliminate floor reactions from moment equation.

Sum of moments about lower end = 0.

Take clockwise moments positive. The weight W acts at the midpoint; its perpendicular distance to the lower end = (L/2) sin θ.

Normal reaction at the wall Nw acts horizontally at the top and produces a moment tending to rotate ladder clockwise; its perpendicular moment arm = L cos θ.

Moment equation: Nw × L cos θ = W × (L/2) sin θ.

Therefore, Nw = (W/2) tan θ.

Friction at wall Fw ≤ μs Nw, and friction at floor Ff ≤ μs Nf.

Horizontal equilibrium gives: Ff = Nw.

Vertical equilibrium gives: Nf = W + Fw.

At impending slip, assume ladder tends to slip down; friction at floor acts up the plane and at wall acts downwards. For limiting case take Fw = μs Nw and Ff = μs Nf.

Substitute Nw from moment equation into horizontal equilibrium:

Ff = (W/2) tan θ.

But Ff = μs Nf = μs (W + Fw).

Also Fw = μs Nw = μs (W/2) tan θ.

Thus Nf = W + μs (W/2) tan θ.

Therefore, (W/2) tan θ = μs [ W + μs (W/2) tan θ ].

Divide through by W and rearrange:

(1/2) tan θ = μs + (μs^2 / 2) tan θ.

Collect terms containing tan θ:

tan θ [ (1/2) - (μs^2 / 2) ] = μs.

tan θ = 2 μs / (1 - μs^2).

Hence the minimum angle θmin is given by:

θmin = arctan [ 2 μs / (1 - μs^2) ].

This expression assumes μs < 1. If μs is large, the ladder will not slip for any practical θ.

Example 5 - Wedge and block (equilibrium on a wedge)

Problem. A block of weight W rests on the rough inclined face of a fixed wedge of angle α. Coefficient of static friction between block and wedge is μs. Determine the range of α for which the block remains at rest.

Sol.

Resolve weight into components perpendicular and parallel to the wedge face.

Normal reaction N = W cos α.

Component tending to slide down the face = W sin α.

Maximum static friction up the face = μs N = μs W cos α.

If W sin α ≤ μs W cos α the block remains at rest (no sliding downwards).

Therefore, tan α ≤ μs.

If tan α < μs the block stays at rest; if tan α = μs it is at the limit of motion; if tan α > μs it will slide downwards.

Common tips for solving friction problems

• Always draw a free-body diagram showing normal, friction and applied forces. Label directions of possible motion; assume a direction for friction opposite that motion.
• Use equilibrium of forces and moments together with the inequality f ≤ μs N.
• For impending motion, replace inequality by equality f = μs N and solve for the unknown.
• Be careful about vertical components that change the normal reaction (e.g., pushing or pulling at an angle).
• Check limiting conditions: e.g., normal reaction must remain non-negative. If the normal goes to zero, the contact is lost and the problem changes.

Applications and remarks

• Dry friction is used beneficially: brakes, clutches, footwear tread, and clamps rely on controlled friction.
• Dry friction leads to energy dissipation as heat when sliding occurs; this is important in wear and lubrication design.
• In structural and mechanical problems, accurate estimation of coefficients μs and μk from experiments or standards is necessary for safe design.
• For engineering problems involving repeated or variable motion, consider whether static or kinetic friction is the appropriate model.

Summary

Dry friction resists relative motion and is governed by simple laws that relate frictional force to the normal reaction via coefficients of friction. Static friction prevents motion up to a limiting value fmax = μs N, and kinetic friction during sliding is f = μk N. The angle of repose gives a direct measure of limiting friction. A systematic free-body diagram, equilibrium equations, and correct use of f ≤ μs N will solve most examination-level problems.