EAS208_HW04_092210_Solutions

EAS208_HW04_092210_Solutions - EAS 208 (Fall 2010) Homework...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 6
Problem 14–43 The boat is moving at 450 kg 10 m s when its engine is shut down. The magnitude of the
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 27

EAS208_HW04_092210_Solutions - EAS 208 (Fall 2010) Homework...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online