# HW 08 - Fa14 - Statics and Gravitation

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HW 08 - fa14 - statics and gravitation:  Ch. 12: 3, 6, 12, 25, 28, 37, 48; Ch. 13: 7, 8, 13, 16, 35, 40, 57. Chapter 12, Problem 3:    In the Figure, a uniform  sphere of mass 0.85 mkg   and radius 4.2 rcm  is held in place by a massless rope attached to a  frictionless wall a distance 8.0  Lcm  above the center of the sphere. Find (a) the tension in the rope and (b) the force on the sphere from the wall.   Notice our introduction of the placeholder-variable   , in addition to what’s already in the text. This helps us compute the  x -component of the tension from the  y -component, because tan//  xy TTrL      , so, 222222 0; 0; (tan)()(/)1;  xxxyyxy  FRTFTmgTTTmgmgmgrL                  (1.1) In (1.1),  x  R  is the reaction-force from the wall, so, plugging in numbers, 2 4.2 (0.85)(9.81)( cmm s Tkg   8.0 cm 2 24.2 )19.418; tan(/)(0.85)(9.81) cmm xx s  NRTmgmgrLkg          8.0 cm 4.378;  N   (1.2) chapter 12, problem 6:    A scaffold of mass 60 mkg   and length 5.0  Lm   is supported in a horizontal position by a vertical cable at each end. A window washer of mass 80  Mkg    stands at a point 1.5  xm   from one end. What is the tension in (a) the nearer cable and (b) the farther cable?  The weight of the window-washer exerts a torque about one end of the scaffold (which is basically a uniform stick (but 2-dimensional)) 2 1.511225.02 0((/))(805260)(9.819.74); mm L farfar m s  xMgmgLTTxLMmgkgN kg                 (1.3)  Newton’s 2 nd  Law in the y-direction, meanwhile, yields, 2 0()529.748()(143.66;40)(9.81) m ynearfarnearfar  s  FTTmMgTmMgTN kgN              (1.4) The symbolic expression for the “near” force is: 12 ((1))  xnear  L TmMg     . Chapter 12, problem 12:    In the Figure, trying to get his car out of mud, a man ties one end of a rope around the  front bumper and the other end tightly around a utility  pole 18  Lm   away. He then pushes sideways on the rope at its midpoint with a force of  550  FN    , displacing the center of the rope 0.30  ym   , but the car barely moves. What is the magnitude of the force on the car from the rope? (The rope stretches somewhat.) Again, we use the equation tan//  yx TTyx      , where 12  xL  , immediately yielding, 1822(0.30) 0; 0cotcot15005506 m L yyyxxxxxy ym  FFTTFFTRRTTFNN  F                      (1.5)  ããchapter 12, problem 25:    In the Figure, what magnitude of (constant) force applied horizontally at the axle of the wheel is necessary to raise the wheel over an obstacle of height  3.00 hcm  ? The wheel’s radius is 6.00 rcm   , and its mass is 0.800 mkg   . Assume the wheel mounts the corner at constant velocity. This is difficult to maintain physically, but necessary to solve the problem. Drawing a schematic, (1.6) Torque balance, x-force balance, and y-force balance, 0sinsin; 0; 0;  xyxy  RRxRryRrxxyy  RrRrFFRFRmg                       (1.7) Trig functions are, 22 sin(90)cos()/; sinsin(90(90))sin()/;  xxyxx  RrRrRrRrRr  rrhrrhr                    (1.8) Combining (1.7) and (1.8), and solving for  F  , we initially have cossin  xx  RrRr   Frmgr       (in which we used the machinations of (1.8)), which is readily rearranged into, 2 222222 ()/()cot()/(0.800)(9.81)(6.00)(3.00);(3.13.9)500  x  Rr m s rrhrmgrrh Fmgmg rhrrhkgcmcm N cm             (1.9) Fun plotting exercise suggested by the solutions manual:   The applied force here is about 1.73 times the weight of the wheel. If the height is increased, the force that must be applied also goes up. Next we plot  F  / mg   as a function of the ratio / hr  . The required force increases rapidly as /1 hr    .   You should get in the habit of plotting things—see if you can reproduce this above plot using my result (1.9). chapter 12, problem 28:    In Fig. 12-43, suppose the length L of the uniform bar is 3.00 m and its weight is 200 N. Also, let the block’s weight W = 300 N and the angle 30.0      . The wire can withstand a maximum tension of 500 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x (pretending the string doesn’t break), what are the (b) horizontal and (c) vertical components of the  force on the bar from the hinge at A?  (a) the torque-balance about point-A determines the max distance the block can stand on the beam,   max 1,maxmax2solve for xmaxmax 0sin()sin()sin90()sin90sin/2(500N)sin30.0(200N)/23.00 m1.50m;300N iwireblockrodCMiiibar b rFLTxWLW TW  xLW                                 (1.10) (b) The x-reaction-force F x  is due to , 0  xi  F     , in which there are two forces: the (negative) x-component of tension, and the (positive) reaction-force, solve for F,maxmax 0cos=cos=433N;  x  xixxxx  FTFTFFT               (1.11) (c) the y-reaction force F y  is due to , 0  yi  F     ; unlike the x-direction (1.11), there are four forces in this  balance, solve for F,maxmax 0sin=+sin=250N;  x  yiyyblockbarybyb  FFTWWFTWWFWWT                 (1.12)  chapter 12, problem 28:    In the Figure, take 3.00  Lm  and its weight is 200 bar  WN   . Also, let the block’s weight 300 WN   and the angle 30.0      . The wire can withstand a maximum tension of  max 500 TN   . (a) What is the maximum  possible distance x before the wire breaks? With the block placed at this maximum x (pretending the  string doesn’t break), what are the (b) horizontal and (c) vertical components of the force on the bar  from the hinge at A?  (a) the torque-balance about point-A determines the max distance the block can stand on the beam,   max 1,maxmax2solve for xmaxmax 0sin()sin()sin90()sin90sin/2(500N)sin30.0(200N)/23.00 m1.50m;300N iwireblockrodCMiiibar b rFLTxWLW TW  xLW                                 (1.13) (b) The x-reaction-force F x  is due to , 0  xi  F     , in which there are two forces: the (negative) x-component of tension, and the (positive) reaction-force, ,maxmax 0coscos500cos30.0433;  x  solveforF  xixxxx  FTFTFFTNN                   (1.14) (c) the y-reaction force F y  is due to , 0  yi  F     ; unlike the x-direction (1.11), there are four forces in this  balance, ,max max 0sinsin300200500sin30.0250;  x  yiyyblockbaryb solveforF  ybar   FFTWWFTWW  FWWTNNNN                        (1.15) chapter 12, problem 37:    In the Figure, a uniform plank, with a length 6.10  Lm   and a weight of  445 WNmg     , rests on the ground and against a frictionless roller at the top of a wall of height  3.05 hm  . The plank remains in equilibrium for any value of 70       but slips if 70      . Find the coefficient of static  friction between the plank and the ground.  We choose the rotation axis to be such that the reaction force ˆˆ  xy  RRiRj      as illustrated in the schematic makes zero contribution to the torque, ; (1.16)

Jul 23, 2017

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