Friction and Wedges

Introduction

Friction always acts tangent to the surface an object is sliding on.

Types of Friction Problems:

1. No impending motion

   1. Equations:

      1. Force of Static Friction: $F_{s} = \mu_{s}N$, where $\ \mu_{s} =$ coefficient of static friction

      2. Angle of Static Friction: $\Phi_{s} = \tan^{- 1}\frac{F_{s}}{N} = \tan^{- 1}\frac{\mu_{s}N}{N} = \tan^{- 1}\mu_{s}$

   2. Solved using Equilibrium Equations

   3. Friction is obtained using $F_{s} \leq \mu_{s}N$

2. Impending Motion

   

   1. Equations:

   2. Force of Static Friction: $F_{k} = \mu_{k}N$, where $\mu_{k} =$ coefficient of kinetic friction

   3. Solve using:

      1. $F_{s} = \mu_{s}N$ for impending motion

      2. $F_{k} = \mu_{s}N$ for slipping motion

Wedges

Wedges are simple machines that transform an applied load into a large load, or can adjust heavy loads.

Steps for Solving

• Draw a FBD and label friction forces as unknowns.

• Solve for equilibrium equations, knowing that normal force is perpendicular to the surface and weight acts straight down.

• You cannot always assume that $F = \mu N$ you wouldn’t know if the object slips or tips. If it is given, then you can solve the equation.

Example 1: Problem 8-5

The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required to move the car if the back brakes are locked, and the front wheels are free to roll. Take $\mu_{s} = 0.3$.

$\circlearrowleft\Sigma M_{A} = 0;$

$0 = - F( - 0.3m \cdot cos(30) + 0.75m \cdot sin(30)) - 19.62kN \cdot 1m + N_{B} \cdot 2.5m$

$\Rightarrow N_{B} = 0.046F + 7.848kN$

$f_{b} = \mu_{s}N_{B} \Rightarrow f_{B} = 0.3(0.046F + 7.848kN)$

$\rightarrow \Sigma F_{x} = 0;$

$- Fcos30 + 0.3(0.046F + 7.848kN) \Rightarrow F = 2.76kN = 0$

Example 2: Problem 8-58

The blocks each have a weight of 50 lb. If the coefficient of static friction at A is

$\mu_{s} = 0.2$ and between each block ${\mu'}_{s} = 0.4$, determine how many blocks can be stacked as shown before they begin to topple.

$\Phi = \tan^{- 1}({{\mu'}_{s})}_{}$

$\Phi = \tan^{- 1}(0.4) \Rightarrow \Phi = 21.8$

Since $\Phi < 21.8$, an unlimited number of blocks can be stacked.