8 Friction and Mechanical Applications

Learn how to model dry friction, determine friction directions, and analyze equilibrium in blocks, wedges, and belt–pulley systems.

The dry- model

acts at a contact and is parallel to the surface. It resists relative sliding, or the tendency for relative sliding, between the contacting bodies. It can enable wedges and belts to transmit forces, while also limiting their motion.

In the Coulomb dry- model, the contact force has a normal component perpendicular to the surfaces and a component tangent to them. The normal force is denoted by $N$ , and the force by $F$ . The coefficients of depend on the pair of surfaces and are dimensionless.

Static and

prevents relative sliding and adjusts to the amount required, up to its limiting value:

0 \leq F_s \leq \mu_s N

Here, $\mu_s$ is the coefficient of . At , reaches its maximum, so $F_s=\mu_sN$ . Do not assume the maximum value applies unless slip is impending; otherwise, determine the required for equilibrium and check that it does not exceed the limit.

acts when surfaces are sliding relative to each other. In the simplified model, its magnitude is

F_k=\mu_kN

Here, $\mu_k$ is the coefficient of , usually less than or equal to $\mu_s$ . opposes relative motion or impending relative motion at the contact, not necessarily the overall motion of the body. For example, on a driven wheel can point in the same direction as the wheel’s motion.

Equilibrium and free-body diagrams

A (FBD) shows the weight, applied forces, support reactions, and contact forces acting on the selected body. At a rough contact, draw the normal force perpendicular to the surface and parallel to it. When two bodies touch, the forces they exert on one another are equal and opposite, but each force belongs on the of the body it acts on.

For a rigid body in planar static equilibrium, apply force balance in both coordinate directions and moment balance about a point $O$ :

\sum F_x=0, \qquad \sum F_y=0, \qquad \sum M_O=0

To determine a direction, imagine the relative motion that would occur if vanished, or use the stated impending motion. opposes the resulting relative slip at that contact. If an assumed direction yields a negative answer, the actual direction is opposite. For multiple bodies, draw a separate diagram for each body and determine the direction separately at each contact.

Block on a horizontal floor

A $10\,\text{kg}$ block rests on a level floor and is acted on by a horizontal force of $30\,\text{N}$ . Given $\mu_s=0.40$ , and assuming there are no other vertical forces, the normal force is $N=mg=98.1\,\text{N}$ . The maximum is

F_{s,\max}=\mu_sN=(0.40)(98.1)=39.2\,\text{N}

Because the applied force is less than this limit, the block can remain at rest. The actual force is $30\,\text{N}$ , opposite the applied force, rather than $39.2\,\text{N}$ . If the applied force is increased beyond $39.2\,\text{N}$ , static equilibrium is no longer possible.

Blocks on inclined surfaces

For a block on a plane inclined at angle $\theta$ above the horizontal, choose axes parallel and perpendicular to the plane. Resolve its weight, $W=mg$ , into the components

W_{\parallel}=mg\sin\theta, \qquad W_{\perp}=mg\cos\theta

If no other force has a component perpendicular to the plane, the normal force is $N=mg\cos\theta$ . For a block at rest with no applied force along the plane, must balance the downslope component: $F_s=mg\sin\theta$ , provided this required is no greater than $\mu_sN$ .

At impending downslope slip, the downslope component equals the limiting :

mg\sin\theta=\mu_smg\cos\theta, \qquad \mu_s=\tan\theta

This result applies to the stated simple setup. Additional forces can change both the normal force and the required.

Wedges

A wedge is a tapered body that converts an input force into forces that separate or lift other bodies. Its inclined faces create contact normal forces, while on a rough face resists relative sliding at that interface. A wedge can be driven in, withdrawn, or held in place; these different motion tendencies can reverse the directions.

Analyze a wedge by drawing a for the wedge and for each body it contacts. Include the input force, weight where relevant, and the normal and forces at each contact. At a contact with , use $F_s=\mu_sN$ ; when sliding is specified, use $F_k=\mu_kN$ . Resolve forces on angled faces into convenient coordinate directions and apply equilibrium to each body.

Determine the relative tendency to slip separately at every interface; does not necessarily point in the same direction at all contacts. Narrow wedges can provide substantial force amplification. can resist driving a wedge or help it hold position.

Belts on pulleys

A belt wrapped around a rough cylindrical pulley can sustain different tensions on its two sides. distributed along the contact arc supports this tension difference. For a flexible belt about to slip on a flat pulley, the limiting is

\frac{T_{\text{tight}}}{T_{\text{slack}}}=e^{\mu_s\beta}

Here, $T_{\text{tight}}$ and $T_{\text{slack}}$ are the larger and smaller belt tensions, $\mu_s$ is the coefficient of , and the wrap angle $\beta$ is measured in radians. For no slip, the tension ratio must not exceed the limiting value. The relation assumes a flexible belt and ; for a different impending direction, identify which side becomes the tight side.

Half-wrap example

For $\mu_s=0.30$ and a half-wrap of $\beta=\pi$ radians, the tension ratio at is

\frac{T_{\text{tight}}}{T_{\text{slack}}}=e^{0.30\pi}\approx 2.57

If the slack-side tension is $100\,\text{N}$ , the largest tight-side tension before slip is approximately $257\,\text{N}$ . A smaller tension ratio can be supported without reaching the limit.

The dry-friction model

Static and kinetic friction