EAS208_HW04_092210_Solutions

# EAS208_HW04_092210_Solutions - EAS 208(Fall 2010 Homework#4...

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EAS 208 (Fall 2010) Homework #4 – Solutions Problem 13–164 Relative to the earth-fixed coordinate system, the disk rotates about the fixed point with a constant angular velocity of O 10 rad s . What is the acceleration of point A relative to point B at the instant shown? Answer: The acceleration in Cartesian coordinates at points and A B are, respectively: 2 22 2 10 200 A ft rad Rf t ss ω =− a jj j 2 2 10 200 B ft rad t ai i i Therefore the acceleration of point relative to point A B at the instant shown is: () 2 // / 200 200 A B AB A B ft s =+ =−⇒ = aaa a aa a i j 1

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Problem 14–7 The coefficient of kinetic friction between the 14 box and the inclined surface is kg 0.1 k μ = . The velocity of the box is zero when it is subjected to a constant horizontal force 20 F N = . What is the velocity of the box two seconds later? Answer: Plot the free-body diagram of the box: Analyze the motion in the direction parallel to the inclined surface and normal to it: sin 20 cos20 sin 20 oo x F m a W fF m a m g m a Σ = ⇒− −= 0 sin 20 0 cos 20 sin 20 0 y F W NF m g Σ=⇒ −− =⇒ = 2
Assume that the box slips, thus the friction force f is: 0.1 k f NN μ = = Solving the system of equations we find that: 2 1.04 sin 20 cos20 46.97 0.1 18.79 14 129.06 6.84 0 sin 20 0 135.9 oo m a mg f F ma N a s N mg N F = −− = = ⎪⎪ ⎨⎨ = = = and 0.1 135.9 13.59N k fN f == × = Since the assumption that the box is sliding downwards is valid. 0 a > Using kinematics: 2 1.04 2 2.08 m va t s v s s = m down the inclined surface 3

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Problem 14–26 The robot manipulator is programmed so that it is stationary at 0 t = and the components of acceleration of A are 2 400 0.8 xx mm av s =− and 2 200 0.4 yy mm s from 0 t = to , where 2 t = s x v and are components of the velocity in y v mm s . The y axis points upward. What are the x and components of the total force exerted by the jaws of the manipulator on the widget y 2 kg A at 1 ts = ? Answer: Plot the free-body diagram of the box: 4
The acceleration in the x direction is: 400 0.8 x x x dv av dt ==− Separate variables and integrate to obtain: () ( 00 11 1 ln 400 0.8 ln 400 400 0.8 400 0.8 0.8 x v t xx x dt dv dt dv t v vv =⇒ = = −− ∫∫ ) 400 0.8 ln 0.8 0.8 400 400 0.8 400 0.8 0.8 ln 400 400 x v tt te e e ⎛⎞ ⎜⎟ ⎝⎠ ⇒− = = = 0.8 0.8 400 1 500 1 0.8 ve v e ⇒= ( ) mm s At , the velocity at the 1 t = s x direction is: 275.3 x v = ( ) mm s The acceleration in the y direction is: 200 0.4 y yy dv dt Separate variables and integrate to obtain: 1 ln 200 0.4 ln 200 200 0.4 200 0.4 0.4 y v t y dt dv dt dv t v = = 200 0.4 ln 200 0.4 0.4 200 0.4 200 0.4 0.4 ln 200 200 y v e e = = = 0.4 0.4 200 15 0 0 1 0.4 v e ( ) mm s At , the velocity at the 1 t = s y direction is: 164.8 y v = ( ) mm s 5

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At , the acceleration at the 1 t = s x and y direction are, respectively: 2 400 0.8 275.3 179.76 x mm a s =− × = 2 200 0.4 164.8 134.08 y mm a s × = The x and components of the total force exerted by the jaws of the manipulator on the widget y 2 kg A at are, respectively: 1 t = s 2 2 0.180 0.360 xx x x m Fm a F k g F N s Σ= ⇒ =× ⇒ = 22 2 9.81 2 0.134 19.89 yy y y y y mm F ma F mg ma F kg kg F N ss Σ= ⇒ − = ⇒ =× 6
Problem 14–43 The boat is moving at 450 kg 10 m s when its engine is shut down. The magnitude of the

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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_HW04_092210_Solutions - EAS 208(Fall 2010 Homework#4...

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