p0306 - 3.6. CHAPTER 3, PROBLEM 6 151 3.6 Chapter 3,...

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Unformatted text preview: 3.6. CHAPTER 3, PROBLEM 6 151 3.6 Chapter 3, Problem 6 Problem: A block of mass m is pulled up a frictionless incline by a force F as shown. The incline lies at an angle φ to the horizontal and the force is at an angle ψ to the incline. Assume the block remains in contact with the incline. (a) Determine the reaction force of the incline on the block, N. Express your answer in terms of m, φ, ψ , F = |F|, and gravitational acceleration, g . HINT: A coordinate-system rotation greatly simplifies this problem. (b) What is the block’s acceleration, a, in the limiting case N = 0? HINT: The trigonometric relationship cos(φ + ψ ) = cos φ cos ψ − sin φ sin ψ will help simplify your answer. . ... . . . .. ... .... .... ... ... .... ..... . .... .... . .. . . .... . .... . ... . .. . . .. . ..... . . .. . ... ................ . .. ...... .. ........ .. ................. ................... . . . .. .. .. ... . . . . .. .. ....... . .................... . . . ...... .. ...... . .......................... . . .. .. .... ....... . . . . . . ... . . ... . . ... ..... .. .. ....... . .. . . ........ .. ... ... ............. .. ................. . ................ .. .. ........................ . . .... .. ..... ............... . .......................... ....................... . . .. . . . ............ . .. . ........................... ...... . ..... .. . . ................................ . . .... . . ... . ...... ............................... ............ . ........................................... .................. .............................................. ......................... .. ............................................. .......................... . ...... ................... . ... ... . .. . ......................................................... . ....................................... . . .. . .......................................................... . ......... . .. ... ... .. ......................................... .. . ... .. . .............................................. .. . .. . ... . ........................................ . . .. ............ . . ...... . .. .. ............................ .. . .. . ........................... . ... ..... .. ... ............... . . . ... ... . .. .. ........ . . . .. .. . .. .............. . ............... ... . . .................. . ...................... .. .. .. . .. . .. . . ..... . ........ . ................. . .. . .................. ................... . . .. . . .... .. . .. ......... ............... .. .. .................. ................. . . .. .. . .. . ..... .. .. .......... .. ............. .. .. ................... . .. . ........................... . .... .. ................. . ... ......................... .. . ..... ........... .. ........................................................................................ . ................. . .... ... ......................... ....................................................... . . . . . . ... . z z F ψ x m g = −g k N φ x Solution: Because the motion occurs parallel to the incline, it is convenient to first transform to the x z coordinate system. We do this by rotating the axes through an angle φ. The unit vectors i and k for the xz system can be expressed in terms of unit vectors i and k for the x and z system as follows. i = i cos φ − k sin φ (a) The three forces acting are F = F i cos ψ + k sin ψ N = Nk mg = −mg k = −mg i sin φ + k cos φ So, Newton’s Second Law in the x z Cartesian coordinate system becomes m d2 x dt2 d2 z m2 dt = F cos ψ − mg sin φ = F sin ψ − mg cos φ + N and k = i sin φ + k cos φ Since there is no motion normal to the incline, necessarily d2 z /dt2 = 0, so that the magnitude of the normal force is N = mg cos φ − F sin ψ (b) In the limiting case N = 0, we have the following relation between F and mg , F = mg cos φ sin ψ 152 CHAPTER 3. FORCE AND ACCELERATION So, the acceleration parallel to the incline, ax = d2 x /dt2 , is given by max = mg cos φ cos ψ − mg sin φ sin ψ =⇒ ax = g cos φ cos ψ − sin φ sin ψ sin ψ Finally, taking advantage of a familiar trigonometric identity, the acceleration of the block is a=g cos(φ + ψ ) i sin ψ ...
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This note was uploaded on 10/15/2010 for the course AME 301 at USC.

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