EAS208_HW08_111510_Solutions

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

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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 Wl b = 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 xA B W Fm a Ff f a g Σ= ⇒− − = x (1) The equation of motion in the y -direction is (no motion in the y -direction): 0 y y yy a AB W ABW Σ = ⇒+−= ⇒+= (2) From the Coulomb friction equation: k f N μ = ( ) k y y f fA += + B k f fW + = (3) From Eq. (1) and Eq. (3): W Ff f a g −−= x kx W FW a g −= xk F ag W ⎛⎞ =− ⎜⎟ ⎝⎠ 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
(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 6 0 2 82 81 41 4 GA B y MF f f A B Σ = −−−+ 0 y = 32 22 14 yA B y B Ff f A = +++ (5) From Eq. (2), Eq. (3) and Eq. (5): 32 14 B y B f A = 32 2 14 yk y B FW W B μ =+ + 32 1 28 2 B FWW + 32 1 30 0.1 230 230 28 2 y B lb + 172.3 y B lb = From Eq. (2): 230 172.3 yy y y y ABW AW B A l b +=⇒= −⇒= 57.7 y Al b = 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
Calculate the position of the center of mass of the object (using point B as reference): 2 BDE total G BDE l md m = 2 BDE BDE G total ml d m =⋅ 30 . 6 0.18 232 G dm = + The equation of motion in the x -direction is: x total Gx x total Gx Fm a Dm a Σ= ⇒ = (1) The equation of motion in the y -direction is: y total Gy y total total Gy a g m a ⇒ − = (2) Take moments around fixed point : D DD M I α Σ = (3) where () ( ) 22 2 2 2 0.4 0.6 0.2 0 . 4 330 . 4 7 12 12 2 D mm m I kg kg m kg kg kg m ⎛⎞ =+ ++ = ⎜⎟ ⎝⎠ Thus: M I Σ = 0.4 0.1 ABC BDE D mg I +⋅ = 0.4 0.1 ABC BDE D I = 2 2 9.81 0.4 3 9.81 0.1 23 0.47 rad s ⋅⋅ + == 5

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From kinematics of rigid body for fixed-axis rotation: 2 GD G G D D rr αω =+ × aa r r r ()( ) () () ( ) ( ) 2 0 23 0.40 0.18 5 0.40 0.18 G ×− ak i r i 5.5 5.06 G = ai j ( ) 2 m s (4) Therefore, from Eq. (4): 5.5 Gx = ( ) 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 Dm a D N == = ( ) 5 5.05 9.81 23.8 y total Gy y a g D N = + = 6
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 of the velocities of their centers when they have rolled a distance .

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## This document was uploaded on 10/27/2011 for the course EAS 208 at SUNY Buffalo.

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EAS208_HW08_111510_Solutions - EAS 208 (Fall 2010) Homework...

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