Homework 11 Solutions - homework 11 ALIBHAI ZAHID Due 4:00...

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homework 11 – ALIBHAI, ZAHID – Due: Apr 11 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Nobody at the playground wants to play with an obnoxious boy, so he fashions a seesaw as shown so he can play by himself. Explain how this is done. 1. The angular velocity of the boy is can- celled with that of the board. 2. The weight of the boy is balanced with an unknown heavy metal. 3. The weight of the boy is balanced by the weight of the board. correct 4. The fulcrum is very far from the boy. Explanation: The weight of the boy is counterbalanced by the weight of the board, which can be consid- ered to be concentrated at its center of gravity on the opposite side of the fulcrum. He is in balance when his weight multiplied by his dis- tance from the fulcrum is equal to the weight of the entire board multiplied by the distance between the fulcrum and the midpoint (center of gravity) of the board. keywords: torque Question 2 part 1 of 1 10 points Consider a solid sphere of radius R and mass m held against a wall by a string being pulled at an angle θ . m g P F θ R Determine the torque equation about the point P . 1. R m g = R F 2. R m g = 2 sin θ R F 3. R m g = (2 sin θ ) R F 4. R m g = (1 + cos θ ) R F 5. R m g = (2 cos θ ) R F 6. R m g = 2 cos θ R F 7. R m g = (1 + sin θ ) R F correct Explanation: m g P F θ A θ About P, the clockwise torque is, τ cw = AP × F , where AP = R ( 1 + sin θ ) and the clockwise torque, τ ccw = R m g . Therefore R m g = (1 + sin θ ) R F .
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homework 11 – ALIBHAI, ZAHID – Due: Apr 11 2007, 4:00 am 2 Question 3 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. at the critical point of slipping. 2. stable. 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 4 part 1 of 1 10 points A string provides a horizontal force which acts on a rectangular block of weight 420 N at top right-hand corner as shown in the figure below. 1 m 0 . 8 m F 420 N If the block slides with constant speed, find the tension in the string required to start to tip the block over. Correct answer: 168 N (tolerance ± 1 %). Explanation: Let : F = 168 N , W = 420 N , h = 1 m , and w = 0 . 8 m . h w F W Locate the origin at the bottom left corner of the block. In equilibrium (at constant velocity; i.e. , no acceleration), we have summationdisplay vector F = 0, so summationdisplay F x = F μ N = 0 (1) summationdisplay F y = N − W = 0 , (2) The block is trying to tip over its lower right-hand corner (the fulcrum). The string tension T pulls to the right (clockwise) at a distance h from that corner. The weight W acts down (counter-clockwise) at a distance w 2 from the corner. From the torques summationdisplay vector τ = 0 , so summationdisplay vector τ = 1 2 W w F h = 0 . (3) F h = W w 2
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homework 11 – ALIBHAI, ZAHID – Due: Apr 11 2007, 4:00 am 3 F = W w 2 h = (420 N) (0 . 8 m) 2 (1 m) = 168 N .
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