HW__4_Solutions - Physics 112 - HW #4 Solutions 4-26 [Step...

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- 1 - Physics 112 — HW #4 Solutions Spring 2006 4-26 [Step on it!] Assuming the truck is accelerating to the right ( ), as shown: (a) Forces on Box: Forces on Truck: f 1 N 1 w = m g B B (normal contact force due to Truck) (friction contact force due to Truck) (weight due to Earth) f 2 N 2 N R f R w = m g T T (normal contact force due to Box) (friction contact force due to Box) (friction contact force due to Road) (drag force due to Air and Road) f d (weight due to Earth) (normal contact force due to Road) NOTE CAREFULLY!: * The kinetic (sliding) friction forces f 1 and f 2 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. * N 2 , 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! (b) There are two 3rd Law interaction pairs among the forces shown above: (i) N 1 and N 2 (normal contact interaction between the Box and Truck) (ii) f 1 and f 2 (frictional contact interaction between the Box and Truck) a
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- 2 - 4-33 [Force is a Vector] (a) The smallest force the child could apply to make the total force acting on the cart be in the +x direction would be a force in the ± y direction to balance the y-component of the sum of the forces applied by the adults: F 1y + F 2y = F 1 sin60° - F 2 sin30° = (100 N)(0.866) - (140 N)(0.500) = +17 N
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HW__4_Solutions - Physics 112 - HW #4 Solutions 4-26 [Step...

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