Engineering Mechanics

9.1. INTRODUCTION In the last chapter, we have discussed the principles of friction of various types. Though these principles ha ...

(^150) „„„„„ A Textbook of Engineering Mechanics Fig. 9.2. Fig. 9.1. Ladder friction Consider a ladder AB resting on the rough g ...

Chapter 9 : Applications of Friction „„„„„ 151 Equilibrium of the ladder We know that the maximum force of friction available at ...

(^152) „„„„„ A Textbook of Engineering Mechanics sin 45° = cos 45° = 0.707 Solution. Given: Length of the ladder (l) = 4 μ; Angl ...

Chapter 9 : Applications of Friction „„„„„ 153 Example 9.4. A uniform ladder 3 m long weighs 200 N. It is placed against a wall ...

(^154) „„„„„ A Textbook of Engineering Mechanics Taking moments about A and equating the same, (200 × 1.5 cos 60°) + 1000 × 3 co ...

Chapter 9 : Applications of Friction „„„„„ 155 Dividing both sides by 3 cos α, 1000 + 100 = Fw + Rw tan α 1100 = 99.1 + 385.4 ta ...

(^156) „„„„„ A Textbook of Engineering Mechanics Taking moments of the forces about the block A and equating the same, Rw × l co ...

Chapter 9 : Applications of Friction „„„„„ 157 9.3. WEDGE FRICTION A wedge is, usually, of a triangular or trapezoidal in cross- ...

(^158) „„„„„ A Textbook of Engineering Mechanics Now consider the equilibrium of the body DEFG. We know that the body is in equ ...

Chapter 9 : Applications of Friction „„„„„ 159 Let P = Minimum horizontal force required to raise the block. The example may be ...

(^160) „„„„„ A Textbook of Engineering Mechanics Resolving the forces horizontally, R 1 cos (16.7°) = R 2 sin (10 + 16.7°) = R 2 ...

Chapter 9 : Applications of Friction „„„„„ 161 Solution. Given: Angle of the Wedge (α) = 15°; Weight acting on the body (W) = 10 ...

(^162) „„„„„ A Textbook of Engineering Mechanics and now resolving the forces vertically, R 2 sin (15° + 14°) + 1000 = R 1 cos 1 ...

Chapter 9 : Applications of Friction „„„„„ 163 9.4. SCREW FRICTION The screws, bolts, studs, nuts etc. are widely used in variou ...

(^164) „„„„„ A Textbook of Engineering Mechanics 9.5. RELATION BETWEEN EFFORT AND WEIGHT LIFTED BY A SCREW JACK The screw jack i ...

Chapter 9 : Applications of Friction „„„„„ 165 Solution. Given: Mean diameter of screw jack (d) = 50 mm or radius (r) = 25 mm; P ...

(^166) „„„„„ A Textbook of Engineering Mechanics Solution. Given: No. of threads (n) = 2; Pitch (p) = 4 mm; Mean radius (r) = 25 ...

Chapter 9 : Applications of Friction „„„„„ 167 We know that as there are two threads in a cm, (i.e. n = 2) therefore pitch of th ...

(^168) „„„„„ A Textbook of Engineering Mechanics If there would have been no friction between the screw and the nut, then φ will ...