Engineering Mechanics

Chapter 4 : Parallel Forces and Couples „„„„„ 49 A couple is a pair of forces applied to the key of a lock. Now select some sui ...

(^50) „„„„„ A Textbook of Engineering Mechanics 4.9. ARM OF A COUPLE The perpendicular distance (a), between the lines of action ...

Chapter 4 : Parallel Forces and Couples „„„„„ 51 Fig. 4.14. Example 4.7. A square ABCD has forces acting along its sides as show ...

(^52) „„„„„ A Textbook of Engineering Mechanics Equating the moments (iii) and (iv), PQxPa Qb^22 +×=×()–()× ∴ 22 ()–()Pa Qb x PQ ...

Chapter 4 : Parallel Forces and Couples „„„„„ 53 Let P= Weight supported by the man at A. Q= Weight supported by the man at B. C ...

(^54) „„„„„ A Textbook of Engineering Mechanics A couple consists of (a) two like parallel forces of same magnitude. (b) two li ...

55 Equilibrium of Forces Contents Introduction. Principles of Equilibrium. Methods for the Equilibrium of Coplanar Forces. Anal ...

(^56) „„„„„ A Textbook of Engineering Mechanics 5.2. PRINCIPLES OF EQUILIBRIUM Though there are many principles of equilibrium, ...

Chapter 5 : Equilibrium of Forces „„„„„ 57 ∴∠ CAO= 180° – (∠ AOC + ∠ ACO) = 180° – [(180° – β) + (180° – α)] = 180° – 180° + β – ...

(^58) „„„„„ A Textbook of Engineering Mechanics and 15sin 30 15 0.5 7.76 N. BC sin 75 0.9659 T °× ° Ans Example 5.2. A string AB ...

Chapter 5 : Equilibrium of Forces „„„„„ 59 Again applying Lami’s equation at joint C, 1000 sin 120 sin 120 sin 120 TTBC ==CD °°° ...

(^60) „„„„„ A Textbook of Engineering Mechanics 2 300 sin 30 sin 60 1 TBC ==W °° ...[Q sin (180° – θ) = sin θ] ∴ TBC = 300 sin 3 ...

Chapter 5 : Equilibrium of Forces „„„„„ 61 A rope is connected between two points A and B 120 cm apart at the same level. A loa ...

(^62) „„„„„ A Textbook of Engineering Mechanics Solution. Given : Weight of cylinder = 100 N Fig. 5.12. Let RA= Reaction at A, a ...

Chapter 5 : Equilibrium of Forces „„„„„ 63 If the bottom width of the box is 180 mm, with one side vertical and the other inclin ...

(^64) „„„„„ A Textbook of Engineering Mechanics ∴ 1 200 cos 56.1 200 0.5571 134.2 N. sin 56.1 0.830 R °× ° Ans and 2 200 200 240 ...

Chapter 5 : Equilibrium of Forces „„„„„ 65 Example 5.7. Three cylinders weighting 100 N each and of 80 mm diameter are placed in ...

(^66) „„„„„ A Textbook of Engineering Mechanics Since the triangle OSQ is similar to the triangle OPS, therefore ∠ SOQ is also e ...

Chapter 5 : Equilibrium of Forces „„„„„ 67 or R 4 = 50 + 100 = 150 N Ans. (iii) Pressure exerted by the cylinder B on the wall F ...

(^68) „„„„„ A Textbook of Engineering Mechanics Example 5.9. A uniform rod AB of length 3r remains in equilibrium on a hemispher ...