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Assignment 4 Solutions (Fall 2008)

# Assignment 4 Solutions (Fall 2008) - P207 Fall 2008...

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P207 - Fall 2008 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 = 0 = 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 ? 6 ? 6 b b (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 ? 6 - b b (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 ? 6 ? 6 - b b b 1

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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 ˆ 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 ~ 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 ~ F net = ~ F + ~ F g + ~ F N . We solve for ~ F and get ~ F = - ~ F g - ~ F N . The applied force is purely in the x direction. ( F x = | ~ F | , F y = 0). The force of gravity is purely in the y direction. ( ~ F g = - mg ˆ j ). The normal force has x component -| F N | sin θ and y component | F N | cos θ , where θ = 30 .
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Assignment 4 Solutions (Fall 2008) - P207 Fall 2008...

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