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Download PDF by H. E. Barnacle (Auth.): Mechanics of Automobiles

By H. E. Barnacle (Auth.)

ISBN-10: 0080103030

ISBN-13: 9780080103037

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Extra resources for Mechanics of Automobiles

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Frictional drag = μ . R = 0-2 x 5p x δθ == p . <50 lb. <50 lb in. p . p . 6)\f = 5^ X — loU Moments about i? give Λ150 200 x 8 = J 200 X 8 Λ150 5/7. d0 . 4 . sin Θ + J30 -150 lb in. p. άθ(5 - 4 . /?. /?. p . p . /?. sin fl)Jg? = (20 x 0-866 + 20 x 0-866) + ί 5 x ^ ) - (4 x 0-5 - 4 x 0-5) = 34-64 + 10-45 Hence, p = 200 x 8 = 35-6 lb/in2. 12077 Therefore torque= 5 x 35-06 x —— = 373 lb in. 180 VEHICLE BRAKING AND BRAKES 47 (b) Radial pressure = K. sin 0 and radial thrust on element = 5ΛΓ. (50 and thus frictional drag = 0-2 x 5 .

2a shows the forces acting. Horizontal forces give FR = u . f X 2-5 + W X 4-5 2-5 ... MOTIOM FIG. 2. 2 and fil - 0-0682) = 4-5 X 0-3 X 32-2 11 ; , 4-5 X 0-3 X 32-2 . 2b shows the forces acting. Note with braking, the frictional force opposes the motion and inertia force is in the same direction as motion. f Horizontal forces give FR = μ . 3. A motor car weighs 2800 lb and has a wheel base of 100 in. The centre of gravity is 42 in. behind the front axle and 31 in. above ground level. h. in a distance of 85 ft.

2, the width of the lining be t, and the drum radius be r. f and so, frictional drag of drum on shoe (1) = μ . i = μ■. p1. rod . t 44 MECHANICS OF AUTOMOBILES and on shoe (2) = pR2 = μ . p2 . rod . t. The frictional torque on drum due to elements = μ . Rj . r + μ . r. 5) For equilibrium of the shoes, moments of forces about Qx and Q2 give, if OQx = OQ2 = L, = /?!. r . / . L . px . r . i(r — L . 6) P 2 (x+7) = \ p2. r . / . L . p2 . r . î(r - L . 7) may be evaluated for a given brake provided an assumption is made concerning the variation of px and p2 with 0, and usually these are assumed to be either constant or proportional to sin Θ.

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Mechanics of Automobiles by H. E. Barnacle (Auth.)


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