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Dynamics_Part31

# Dynamics_Part31 -

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Unformatted text preview: Problem Set 4: Problem 3. Problem: A ramp is used to unload a crate of mass m from the top of a recreational vehicle. The top of the ramp is a distance h above the ground, its length is L, and it is inclined to the horizontal at an angle . The ramp has been manufactured so that its sliding-friction coefficient increases linearly from top to bottom according to = o (1 + /L), where is distance along the ramp. If the crate starts from rest, what is its speed when it reaches the bottom of the ramp? Express your answer in terms of o , h, and g, the acceleration of gravity. Solution: To solve, we use the Principle of Work and Energy for the crate's motion along the ramp. This will establish its speed when it reaches the bottom of the ramp. The forces acting on the crate when it's on the ramp are as shown below. Clearly, since the crate is not accelerating in the direction normal to the chute, N = mg cos . Thus, the magnitude of the friction force is mg cos , and it opposes the motion. The tangential component of the box's weight accelerates it as it moves down the chute, and its magnitude is mg sin . ... ... ...... .... ... .............. ..... . . .. .................. .. .................... ..... .............................. .............................. ..... .......... . ................................ ....................... . ... ... ...................................................... . .. . . .................. . .................. ................................................. ....... . . ....................................................... .... ....... .................. ... . . . .. . .. . ..... .... . .................................... . ....................................... ........................... ..................... .. ........... . ... .. . .... ............... . ..... . ................. . .. .. ....... .. ....... .. ............... .. ................. . . ........ .... . ............... . ... ............... . .. . .............. . ............. .. .............. . .. .. ...... ......... . ........... . ............. . . .. .... ... . ....... .... . . .. ............... .................. ............... . . . .... .. ........ . . ............ .. .. ................ ..... ... ..... ........ ... . .. ... .......... ................. .............. .............. .. .. . .. . . .. .... ..... . .. ........ ......... ... . . ............. ............... . .. . ..... ... ............. ....... .. ............. . ............ . .... ...... .. ............... . ................ ....... ............ . ... . .......... ... ............... .............. . ... .............. .. .... .. . ..... .... ........... . .............. .............. ............... .............. .. ..... ....... . ................ . ............... . . . . ................ ............... . ............. . . ... . ... . .......... . . .. ........ . . ............. . .................................................. ................................................... . ... ...... ........ . . ............. ... . N mg N Letting denote tangential distance from the top of the ramp, the work done as the crate traverses the ramp is L U1-2 = 0 mg sin - o 1 + L mg cos d 1 2 2L =L =0 = mg sin - o cos + 1 = mgL sin - o cos 1 + 2 3 = mgL sin - o cos 2 From the geometry, we know that h = L sin , wherefore 3 U1-2 = mgh 1 - o cot 2 ...
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