School Work

HW 08 - Fa14 - Statics and Gravitation

Description
gravitation and statics
Categories
Published
of 11
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
  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)
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks