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OLDHW29 - oldhomewk 29 PAPAGEORGE MATT Due 4:00 am Question...

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oldhomewk 29 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 14, sect 1. part 1 of 1 10 points A ladder is leaning against a smooth wall. There is friction between the ladder and the floor, which may hold the ladder in place; the ladder is stable when μ 1 2 tan θ . m g L f 45 h b F N μ = 0 . 4 The ladder will be 1. stable. 2. at the critical point of slipping. 3. unstable. correct Explanation: Stability requires μ 1 2 tan 45 = 0 . 5 . Since μ = 0 . 04 < 0 . 5 , the ladder is unstable. keywords: static equilibrium, instability Question 2, chap 14, sect 3. part 1 of 3 10 points A uniform rod pivoted at one end “point O ” is free to swing in a vertical plane in a gravitational field. However, it is held in equilibrium by a force F at its other end. x y F F x F y W R x R R y θ O Force vectors are drawn to scale. What is the condition for translational equilibrium along the horizontal x direction? 1. R x F x sin θ = 0 2. R x + F x = 0 correct 3. F x = 0 4. R x F x cos θ = 0 5. F x cos θ R x sin θ = 0 Explanation: Using the distances, angles and forces de- picted in the figure, the condition summationdisplay F x = 0 for translational equilibrium in the x direction gives R x + F x = 0 . Question 3, chap 14, sect 3. part 2 of 3 10 points What is the condition for translational equilibrium along the vertical y direction? 1. R y + F y W = 0 correct 2. R y sin θ + F y sin θ W cos θ = 0 3. W R y + F y = 0 4. R y + F y = 0 5. R y F y + W = 0 Explanation:
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oldhomewk 29 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 2 For the equilibrium in the y direction, summationdisplay F y = 0 gives R y + F y W = 0 . Question 4, chap 14, sect 3. part 3 of 3 10 points Taking the origin (point O ) as the pivot point, what is the condition for rotational equilibrium?
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