HW4_solution - P112 Spring 07 Solutions for HW 4 1....

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P112 Spring 07 Solutions for HW 4 1. [unboxed] (a) a rad = v 2 r = (3.00 m/s) 2 14.0 m = 0.643 m/s 2 (b) a tan = 0.500 m/s 2 a = a rad 2 + a tan 2 = (0.643 m/s 2 ) 2 + (0.500 m/s 2 ) 2 = 0.814 m/s 2 tan θ = a tan a rad = 0.500 m/s 2 0.643 m/s 2 = 0.778 θ = 38° a rad a tan a v !
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2. Assume that the truck is accelerating to the right ( ) as shown: Free-body diagram for box: Free-body diagram for truck: A few things to note: The kinetic (sliding) friction forces 1 f r and 2 f r are in the directions shown because the bottom surface of the box is sliding to the left ( ) relative to the surface of the truck with which it is in contact. Remember, the accelerating truck is not an inertial reference frame. 2 N r , the normal contact force acting on the truck due to the box, is not the same thing as the weight of the box, which is the gravitational force acting on the box due to the Earth. These are conceptually different forces for different interactions and must be treated as such! R f r , the friction contact force due to the road is the friction force that the road exerts on the drive tires. The engine of the truck tries to make the drive tires rotate faster in a clockwise fashion. Thus these tires will push back to the
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell University (Engineering School).

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HW4_solution - P112 Spring 07 Solutions for HW 4 1....

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