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EAS208_HW08_111510_Solutions

EAS208_HW08_111510_Solutions - EAS 208(Fall 2010 Homework#8...

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EAS 208 (Fall 2010) Homework #8 – Solutions Problem 18–2 A horizontal force 30 F lb = is applied to the 230 W lb = refrigerator as shown. The coefficient of kinetic friction at A and B is 0.1 k μ = . a) What is the magnitude of the refrigerator’s acceleration? b) What normal forces are exerted on the refrigerator by the floor at A and B ? Answer: Assume that the refrigerator does not tip over (no angular motion). 1
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(a) The equation of motion in the x -direction is: x x A B W F ma F f f a g Σ = = x (1) The equation of motion in the y -direction is (no motion in the y -direction): 0 y y y y y y F ma A B W A B W Σ = + = + = (2) From the Coulomb friction equation: k f N μ = ( ) A B k y y f f A μ + = + B A B k f f W μ + = (3) From Eq. (1) and Eq. (3): A B W F f f a g = x k x W F W a g μ = x k F a g W μ = 2 30 0.1 32.2 230 x ft a s = 2 0.98 x ft a s = (4) The magnitude of the refrigerator’s acceleration is 2 0.98 x ft a s = . 2
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(b) The summation of moments around any point of the refrigerator is zero. Take moments around the center of mass G of the refrigerator: ( ) ( ) ( ) ( ) ( ) 0 60 28 28 28 14 14 G A B y M F f f A B Σ = ⇒ − + 0 y = 32 2 2 14 y A B y B F f f A = + + + (5) From Eq. (2), Eq. (3) and Eq. (5): 32 2 2 14 y A B y B F f f A = + + + ( ) ( ) 32 2 14 y k y B F W W B μ = + + 32 1 28 2 y k B F W W μ = + + 32 1 30 0.1 230 230 28 2 y B lb = + + 172.3 y B lb = From Eq. (2): 230 172.3 y y y y y A B W A W B A lb + = = = 57.7 y A lb = The normal forces exerted on the refrigerator by the floor at and A B are, respectively: 57.7 y A lb = 172.3 y B lb = 3
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Problem 18–22 The object consists of the slender bar 2 kg ABC welded to the slender bar 3 kg BDE . The y -axis is vertical. At the instant shown, the object has a counterclockwise angular velocity of 5 rad s . Determine the components of the force exerted on it by the pin support. Answer: The F.B.D. of the object at the instant shown is: 4
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Calculate the position of the center of mass of the object (using point B as reference): 2 BDE total G BDE l m d m = 2 BDE BDE G total m l d m = 3 0.6 0.18 2 3 2 G d m = = + The equation of motion in the x -direction is: x total Gx x total Gx F m a D m a Σ = = (1) The equation of motion in the y -direction is: y total Gy y total total Gy F m a D m g m a Σ = = (2) Take moments around fixed point : D D D M I α Σ = (3) where ( ) ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 0.4 0.6 0.2 2 2 0.4 3 3 0.47 12 12 2 D m m m I kg kg m kg kg kg m = + + + = Thus: D D M I α Σ = 0.4 0.1 ABC BDE D m g m g I α + = 0.4 0.1 ABC BDE D m g m g I α + = 2 2 9.81 0.4 3 9.81 0.1 23 0.47 rad s α + = = 5
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From kinematics of rigid body for fixed-axis rotation: 2 G D G G D D r r α ω = + × a a r r r ( ) ( ) ( ) ( ) ( ) ( ) 2 0 23 0.40 0.18 5 0.40 0.18 G = + × − a k i r i 5.5 5.06 G = a i j ( ) 2 m s (4) Therefore, from Eq. (4): 5.5 Gx = a i ( ) 2 m s 5.06 Gy = − a j ( ) 2 m s Therefore, from Eq. (1) and Eq. (2), the components of the force exerted on the object by the pin support are: 5 5.5 27.5 x total Gx x D m a D N = = = ( ) ( ) 5 5.05 9.81 23.8 y total Gy y D m a g D N = + = + = 6
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Problem 18–34 A thin ring and a homogeneous circular disk, each of mass m and radius R , are released from rest on an inclined surface. Determine the ratio ring disk v
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