ps4 - P207 - Fall 2007 Solutions Assignment 4 1 Chapter 5,...

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Unformatted text preview: P207 - Fall 2007 Solutions Assignment 4 1 Chapter 5, Problem 13 (a) The net force on the salami is F net = F gravity + T cord to scale where F gravity is the force of gravity and T cord to scale is the force of the scale. Since F net = ma and a = 0, then F net = = F gravity + T cord to scale and therefore T cord to scale =- F gravity = mg = (11 kg)(10 m / s 2 ) = 110 N. salami scale F gravity T cord to scale T cord to ceiling T cord to salami salami scale F gravity T cord to scale T cord to ceiling T cord to salami a63 a54 a63 a54 a98 a98 (b) The tension T in the string is equal to the force of gravity on the salami so that the net force on the salami is zero. The tension in the string is equivalent to the reading on scale. T = mg = 110 N. salami scale F gravity T cord to scale T cord to wall T cord to salami a63 a54 a27 a45 a98 a98 (c) Again, the tension in the string bal- ances the force of gravity. The ten- sion in the string is the same as the reading on the scale. T = mg = 110 N. left salami right salami scale F gravity T F gravity T T T left salami right salami scale F gravity T F gravity T T T a63 a54 a63 a54 a27 a45 a98 a98 a98 1 2. Chapter 5, Problem 20 (a) The net force on the box is F net = ma = F 1 + F unknown . All of the forces, and the motion of the box are in the i direction. We solve for the unknown force F unknown = ma- F 1 = (2 . 0 kg)(10 m/s 2 )- 20 N = 0 (b) F unknown = ma- F 1 = (2 . 0 kg)(20 m/s 2 )- 20 N = 20 N (c) F unknown = ma- F 1 = (2 . 0 kg)(0 m/s 2 )- 20 N =- 20 N (d) F unknown = ma- F 1 = (2 . 0 kg)(- 10 m/s 2 )- 20 N =- 40 N (e) F unknown = ma- F 1 = (2 . 0 kg)(- 20 m/s 2 )- 20 N =- 60 N 3. Chapter 5, Problem 32 (a) In addition to the applied force vector F , there is also the force of gravity acting on the crate, and the normal force of the ramp on the crate. Since the acceleration of the crate is zero, so is the net force on the crate, namely vector F net = vector F + vector F g + vector F N . We solve for vector F and get vector F =- vector F g- vector F N . The applied force is purely in the x direction. ( F x = | vector F | , F y = 0). The force of gravity is purely in the y direction. ( F g =- mg ). The normal force has x component-| F N | sin and y component | F N | cos , where = 30 . Our equation of equilibrium is vector F =- vector F N- vector...
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This note was uploaded on 05/09/2008 for the course PHYS 2207 taught by Professor Liepe, m during the Fall '07 term at Cornell University (Engineering School).

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ps4 - P207 - Fall 2007 Solutions Assignment 4 1 Chapter 5,...

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