Engineering Mechanics

(^488) „„„„„ A Textbook of Engineering Mechanics As discussed above, the applied force must be able to produce an acceleration o ...

Chapter 24 : Laws of Motion „„„„„ 489 Here we shall discuss the following two cases as shown in Fig. 24.1 (a) and (b) : When th ...

(^490) „„„„„ A Textbook of Engineering Mechanics Example 24.11. A body of mass 50 kg is being lifted by a lift in an office. Fin ...

Chapter 24 : Laws of Motion „„„„„ 491 Solution. Given : Mass of elevator (m) = 2500 kg ; Initial velocity (u) = 0 (because it st ...

(^492) „„„„„ A Textbook of Engineering Mechanics Let a 2 = Constant deceleration i.e. retardation We know that final velocity of ...

Chapter 24 : Laws of Motion „„„„„ 493 Solution. Given : Mass of body A (m 1 ) = 80 kg ; Mass of the body B (m 2 ) = 20 kg; Force ...

(^494) „„„„„ A Textbook of Engineering Mechanics and according to D' Alembert’s principle (P 2 – m 2 a = 0) (T – 58.8) – 20 a = ...

Chapter 24 : Laws of Motion „„„„„ 495 Let M = Mass of the gun, V = Velocity of the gun with which it recoils, m = mass of the bu ...

(^496) „„„„„ A Textbook of Engineering Mechanics We also know that v^2 = u^2 – 2as ...(Minus sign due to retardation) 0 = (0.06) ...

Chapter 24 : Laws of Motion „„„„„ 497 Example 24.20. Two men, standing on a floating boat, run in succession, along its length, ...

(^498) „„„„„ A Textbook of Engineering Mechanics Fig. 24.4. Motion on inclined plane. Now consider a body moving downwards on an ...

Chapter 24 : Laws of Motion „„„„„ 499 We also know that the available force (F) 145 = ma = 2000 × a ∴^145 0.0725 m/s^2 2000 α= = ...

(^500) „„„„„ A Textbook of Engineering Mechanics Fig. 24.8. Solution. Given : Grade = 5% or sin α = 0.05 ; Initial velocity (u) ...

Chapter 24 : Laws of Motion „„„„„ 501 Since the force of friction (1930.2 N) is more than the force which will cause slipping (1 ...

(^502) „„„„„ A Textbook of Engineering Mechanics Which of the following statement is wrong? (a) The matter contained in a body ...

Chapter 25 : Motion of Connected Bodies „„„„„ 503 503503 Contents Introduction. Motion of Two Bodies Connected by a String and ...

(^504) „„„„„ A Textbook of Engineering Mechanics Two bodies connected by a string one of which is hanging free and the other ly ...

Chapter 25 : Motion of Connected Bodies „„„„„ 505 g (m 1 – m 2 ) = a (m 1 + m 2 ) ∴ ( 12 ) 12 gm m a mm − = + From equation (vi) ...

(^506) „„„„„ A Textbook of Engineering Mechanics Equating equations (i) and (ii), 150 g – 2 T = 150 a ...(iii) Now consider the ...

Chapter 25 : Motion of Connected Bodies „„„„„ 507 Solution. Given : First mass (m 1 ) = 15 kg ; Second mass (m 2 ) = 6 kg and th ...