ch07 - Chapter 7 Kinetics of Rigid Bodies Undergoing...

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Unformatted text preview: Chapter 7 Kinetics of Rigid Bodies Undergoing Two-dimensional Motion 7.1 Curvilinear Translation 125 7.1.1 GOAL: a) Assuming an engine that can deliver as much power/torque as demanded, calculate the maximum attainable acceleration x and the normal forces at each tire. b) Which tires (front or back) feel the most normal force? GIVEN: The weight of the car is 4100 lbs. The coefficient of friction between the tires and the ground is 0.85. ASSUME: The car doesnt rotate - the motion remains rectilinear. DRAW: FORMULATE EQUATIONS a) Force balance:- mg * + F 1 * + N 1 + N 2 * = ma * * : F 1 = m x (1) * : N 1 + N 2 = mg (2) Moment balance: F 1 (18 in) + N 2 (54 in)- N 1 (54 in) = 0 F 1 + 3 N 2- 3 N 1 = 0 (3) (2) N 1 = mg- N 2 (4) (4) (3) F 1 + 3 N 2- 3 mg- N 2 = 0 F 1 + 6 N 2- 3 mg = 0 N 2 = 3 mg- F 1 6 (5) (5) (4) N 1 = mg- 3 mg- F 1 6 = 3 mg + F 1 6 (6) We know that the maximal tractive force is equal to N T OT AL and since N 1 + N 2 = mg (from (2)) we have F 1 = N 1 + N 2 = mg (7) (7) (1) x = g = 0 . 85(32 . 2 ft/s 2 ) = 27 . 37 ft / s 2 126 (7) (5) N 2 = (3- ) mg 6 = (3- . 85)4100 lb 6 = 1469 . 2 lb (7) (6) N 1 = (3 + ) mg 6 = (3 + 0 . 85)4100 lb 6 = 2630 . 8 lb b) The rear tires feel 79% more normal force than the front tires. 127 7.1.2 GOAL: Determine by what percentage the normal forces change of the vehicle accelerates forward at 0 . 25 m/s 2 GIVEN: Cars mass, length, acceleration and position of mass center. DRAW: FORMULATE EQUATIONS: Force balance: S * + ( N 1 + N 2- mg ) * = m x * * : S = m x (1) * : N 1 + N 2 = mg (2) BAM about G : LN 2- LN 1 + hS = 0 (3) SOLVE: When x = 0, (1) S = 0 (3) N 1 = N 2 Thus, from (2), N 1 = N 2 = mg 2 When x = 0 . 25 g , we have : S = 0 . 25 mg (4) (4) (3) L ( N 2- N 1 ) =- . 25 hmg (5) (2) (5) N 2- ( mg- N 2 ) =- . 25 hmg L 2 N 2 = mg (1- . 25 h L ) 128 N 2 = mg 2 (1- . 25 h L ) (6) (6) (2) N 1 = mg 2 (1 + . 25 h L ) (7) % increase for N 1 : 100(0 . 25 (0 . 7) (1 . 35) ) = 13% % decrease for N 2 : 100(0 . 25 (0 . 7) (1 . 35) ) = 13% The normal force increases at the rear tires and decreases at the front. 129 7.1.3 GOAL: Find N 1 and N 2 that enforce zero rotation on the car. GIVEN: Geometry, dimensions and mass of car. DRAW: FORMULATE EQUATIONS: Well first look at the system when theres no ground present (and no gravity as well). In this situation there are no normal forces on the wheels. Moment balance: Sh = I = . 2 mgh I The response to S is a counter-clockwise rotational acceleration. Now, add N 1 and N 2 , as shown. Force balance: ( N 1 + N 2 ) * + S * = m x * * : S = m x (1) * : N 1 + N 2 = 0 (2) Moment balance: Sh + N 2 L 2- N 1 L 1 = 0 (3) SOLVE: (1),(2),(3) . 2 mgh + N 2 L 2 + N 2 L 1 = 0 N 2 =- . 2 mgh L 1 + L 2 (4) (4) (2) N 1 = . 2 mgh L 1 + L 2 This shows that a positive force must be exerted at the rear wheel and a negative force at the...
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ch07 - Chapter 7 Kinetics of Rigid Bodies Undergoing...

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