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Newton’s 2nd Law:

10 lb. 100 lb. Section 6. Newton’s 2nd Law:. A body with unbalanced forces will accelerate proportional to the unbalanced force and in the direction of the force. S F = ma. Inertial Force. Define inertial force: F i = - ma opposite to the acceleration direction Then

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Newton’s 2nd Law:

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  1. 10 lb. 100 lb. Section 6 Newton’s 2nd Law: A body with unbalanced forces will accelerate proportional to the unbalanced force and in the direction of the force. SF = ma

  2. Inertial Force • Define inertial force: Fi = -ma opposite to the acceleration direction • Then SF + Fi = 0

  3. FBD W Fwind a Fi v Ff Free Body Diagram FB

  4. 5 in 16 in Problem 6-2: An electric hoist is being used to raise crate A as shown. The crate weighs 200 lbs. When the motor is initially powered, it accelerates to 1800 rpm in 0.75 sec. Determine the tension in the cable during this startup.

  5. 24 in 45 in 112 in Problem 6-3: The sports car shown has a wheel base of 112 in and weighs 3400 lbs. The center of gravity has been located as shown. The car has been tested to accelerated from 0 to 60 mph in 8 sec. During this test, determine the road force for each tire.

  6. 1.5 in 650 8 in 2 in 45 psig Problem 6-7: The compressor mechanism shown is running at a constant rate of 600 rpm, cw. The cylinder pressure is 45 psig, and the piston weighs 0.5 lb. The weight of all other links is negligible. For the instant shown, determine the torque required to operate the compressor.

  7. 500 ft 100 Problem 6-24: A curve in a road, has a radius of 500 ft. The road is slightly banked at a 100 angle. During a rain storm, the coefficient of static friction between rubber tires and the road is 0.4. Determine whether it is safe for a 2500 lb car to proceed through the curve at 55 mph.

  8. 4” 3” 900 3” 190 2” 2” Problem 6-29: For the position shown, the shaft is rotating at 400 rpm. Determine the compressive load on the spring. The weights are 0.5 lb each and the weight of the arms is negligible.

  9. Inertial Torque Newton’s 2nd Law also applies to links that encounter angular inertia SM = Ia Define inertial torque: Ti = -Ia Then: SM + Ti = 0

  10. FBD W Fwind a Ti a Fi v Ff FB Free Body Diagram

  11. T 8 in 0.75 in 10 in 1 in Problem 6-31: The grinding disk, and shaft, shown is made of steel. The motor that drives the disk is started and accelerates to its rated speed of 1200 rpm in 1.25 seconds. Determine the torque transmitted to the grinder shaft.

  12. 250 mm C 400 B 700 mm Problem 6-38: Robotic arm BC has a mass of 15 kg and a moment of inertia about its center of gravity of 3.5 kg m2. At the instant shown, arm BC is lowering with an angular velocity of 3 rad/sec and is accelerating at 8 rad/sec2. Determine the torque required to operate joint B, and the reaction forces at that joint. .

  13. Problem 6-46 For the windshield wiper linkage shown, determine the instantaneous torque required to drive the system and the side loads onto the motor shaft. The motor rotates at a constant rate of 45 rpm, counterclockwise. The friction force from the rubber blade on the windshield is shown. The wiper and arm assembly weighs 1.2 lbs, the center of gravity is shown and the mass moment of inertia, relative to an axis at the center of gravity, is 0.4 lb in s2. The weight of all other links is negligible.

  14. 16 in 6 in 2 in 13 in 700 450 1.0 lb 3.5 in 14 in Problem 6-46 (con’t)

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