656 CHAPTER 9 Polar Coordinates; Vectors See Figure 75. We need to find the magnitude of the force v that would cause the wagon to roll down the hill.A force with the same magnitude in the opposite direction of v will keep the wagon from rolling down the hill.The force of gravity is orthogonal to the level ground, so the force of the wagon due to gravity can be represented by the vector F j 100 g = − Determine the vector projection of Fg onto w, which is the force parallel to the hill. The vector w is given by w i j cos20 sin 20 = ° + ° The vector projection of Fg onto w is v F w w w i j i j 100 sin20 cos 20 sin 20 cos20 sin 20 34.2 cos20 sin 20 g 2 2 2 2 ( ) ( ) ( ) ( ) = ⋅ = − ° ° + ° ° + ° = − ° + ° The magnitude of v is 34.2 pounds, so the magnitude of the force required to keep the wagon from rolling down the hill is 34.2 pounds. Solution Figure 75 208 w F g v 6 Compute Work In elementary physics, the work W done by a constant force F in moving an object from a point A to a point B (that is, displacement) is defined as W AB F magnitude of force displacement ( )( ) = = Work is commonly measured in foot-pounds or in newton-meters (joules). In this definition, it is assumed that the force F is applied along the line of motion. If the constant force F is not along the line of motion, but instead is at an angle θ to the direction of the motion, as illustrated in Figure 76, then the work W done by F in the displacement of an object from A to B is defined as W AB F = ⋅ (12) This definition is compatible with the force-times-distance definition, since W AB AB AB AB AB AB AB AB F F F amount of force in the direction of distance projectionof on 2 ( )( ) = = = ⋅ = ⋅ ↑ Use formula (10) Figure 76 A A B F u Figure 77 (a) 308 308 50(cos 308)i 50(sin 308)j x y F (0, 0) (100, 0) (b) iFi 5 50 Computing Work A girl is pulling a wagon with a force of 50 pounds. How much work is done in moving the wagon 100 feet if the handle makes an angle of 30° with the ground? See Figure 77(a). EXAMPLE 7
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