1. Chapter 11 Statics In this chapter we will study the conditions under which a rigid body does not move. This part of Mechanics is called “Statics”. Even though it is rather restricted in scope it is nevertheless a very important topic Engineers. After all, nobody appreciates buildings, bridges, or other structures that move. (11-1)

2. Conditions under which a rigid body does not move 1. The acceleration A cm of the center of mass is zero A cm = 0  dV cm /dt = 0  Velocity of the center of mass V cm is constant. We consider the special case: V cm = 0 2. The angular acceleration  cm of the center of mass is zero  d  cm /dt = 0  the angular velocity  cm of the center of mass is constant. We consider the special case:  cm = 0 (11-2)

3. Equilibrium Condition I Consider a rigid body on which N external forces F 1, F 2, …, F N are applied. The net external force F net = F 1 + F 2 + …+ F N Newton’s second law connects F net with the acceleration of the center of mass A cm : MA cm = F net. In Statics A cm = 0  The net external force must be zero This vector equation can be broken down into three separate equations, one for each coordinate axis: F xnet = 0 or F 1x + F 2x + … + F Nx = 0 F ynet = 0 or F 1y + F 2y + … + F Ny = 0 F znet = 0 or F 1z + F 2z + … + F Nz = 0 (11-3) F1F1 F2F2 FNFN F net = 0

4. Equilibrium Condition II The net external torque on the rigid body of the figure is:  net =  1 +  2 + …+  N  1 = R 1  F 1,  2 = R 2  F 2,... Newton’s second law connects  net with the acceleration of the center of mass  cm : I o  cm =  net. In Statics  cm = 0   net = 0 The net torque by all external forces must vanish (11-4) F1F1 F2F2 FNFN r1r1  net = 0

5. In statics the value of  net does not depend on the choice of the origin So how do we go about choosing the origin O for a statics problem? The answer is that since the choice of O does not affect the result, as a practical matter we choose the origin so as to make the equations simpler. This will become clearer after we work on a few examples. (11-5)

6. Example (11-1) page 301 Consider figure (b) which shows the book resting on the table so that the book’s center of mass lies above the table. In this case F N -mg = 0 because F ynet must be zero. We take the origin to be the edge of the table because when the book starts to fall it will rotate about this point. The net torque  net about O is also zero and the book is at rest. If we perturb slightly the book by lifting it up, F N = 0 and the torque of mg brings it back (11-6)

7. Consider figure (c) which shows the book resting on the table so that the book’s center of mass lies beyond the table. In this case F N -mg = 0 because F ynet must be zero. The net torque  net in this case about O is provided my mg only and it is not zero. Thus the second equilibrium condition is not satisfied and the book will rotate about point O and fall off the table (11-7)

8. ... A B CM mg  A = 0... A B CM mg  B = 0 In statics the extension of the gravitational force must pass through the point of suspension. This is indicated in the two figures. The flat object is suspended first from point A and then from point B. In each case the extension of the gravitational force passes though the point of suspension. In each case the torque of mg is zero and the second equilibrium condition is satisfied. (11-8)

9. ... A B CM mg  A = 0. mg. A CM r  Fig.a Fig.b What happens if we perturb the object from its equilibrium condition as in fig.a? In this case the torque of mg is not zero any more. Instead  = -mgrsin   0 Under the action of mg the object will rotate about A in the clockwise direction till it reaches the position of fig.b. In this position, the torque is zero and the object can be at equilibrium(11-9)

10. Practical method for the determination of the center of mass of a flat object with a complicated shape (such as the map of the US). Suspend the object from a point A (other than the CM). Mark the position of the plumb line. Repeat the process for a second point B. The intersection of the two plumb lines gives the CM. If the process is repeated for a third point C the plumb line passes also from the Center of mass. (11-10) A B CM C

11. Statics Problem Recipe 1. Draw a force diagram 2. Choose a convenient origin O. A good choice is to have one of the unknown forces acting at O 3.Sign of the torque  for each force: + If the force induces clockwise (CW) rotation - If the force induces counter-clockwise (CCW) rotation 4. Equilibrium conditions: F xnet = 0, F ynet = 0,  O = 0 5.Make sure that: numbers of unknowns = number of equations (11-11)

12. Example (11-2) page 303 Find the forces F A and F B on the man’s hands if: m = 5 kg, L = 3 m, = 0.75 m F ynet = F A + F B - mg = 0 eqs.1  net = F B - Lmg = 0 eqs.2 We solve eqs.2 for F B :  F B = Lmg/ = 3  5  9.8/0.75 = 196 N We solve eqs.1 for F A :  F A = mg - F B = 5  9.8 - 196 = -147 N Note: The direction of F A is opposite to that shown in the figure (11-12) y CCW CW

13. Example (11-3) page 304 A crane whose cabin and engine are fixed to earth is used to lift a mass m = 5300 kg. The arm of the crane is supported at its base (point B), by a friction free pivot, and at its top (point A) by a cable. The arm of the cable makes angles  arm = 45  and  T = 32 , respectively, with the horizontal and the arm has a length L = 10 m. The mass m is lifted from a point on the arm at a distance = 0.52 m from point A. Ignore the mass of the arm. Determine the tension T of the cable. (11-13)

14.  net = -mgd 2 + Td 1 = 0  T = mg(d 2 /d 1 ) eqs.1 From triangle ABC:  d 1 = Lsin  eqs.2 Angle  =  arm -  T  = 45  -32  = 13  From triangle BED:  d 2 = (L - )cos  arm eqs.3 We substitute d 1 from eqs.2 and d 2 from eqs.3 into eqs.1  T = mg (L - )cos  arm / Lsin  = 1.55  10 5 N (11-14) d1d1 d2d2 C D E  CCW CW

15. F xnet = F Nx - Tcos  T = 0  F Nx = Tcos  T F Nx = 1.55  10 5  cos32  F Nx = 1.31  10 5 N F ynet = F Ny -mg - Tsin  T = 0  F Ny = mg + Tsin  T F Ny = 5300  9.8 + 1.55  10 5  sin32  = 1.34  10 5 N F N = [F Nx 2 + F Ny 2 ] 1/2 = 1.87  10 5 N (11-15)

16. Example (11-4) page 306 A ladder of length L = 3 m is leaning against the wall making an angle  = 58  with the horizontal. The mass of the ladder is insignificant compared to the mass m = 85 kg of a window washer who is climbing the ladder. The coefficient of static friction between the ladder feet and the floor  s = 0.51. There is no friction between the top of the ladder and the wall. Determine the maximum distance d the man can climb up the ladder so that the ladder starts slipping(11-16)

17. System = ladder + window washer F xnet = F wall – f = 0  F wall = f eqs.1 F ynet = F floor – mg = 0  F floor = mg eqs.2 f =  s F floor =  s mg f =  s mg eqs.3  net = -F wall d 2 + mgd 1 = 0 d 1 = dcos  and d 2 = Lsin   -F wall Lsin  +mg dcos  = 0  d = F wall Lsin  /mgcos  eqs.4 d1d1 d2d2 (11-17) CCW CW

18. d = F wall Lsin  /mgcos  eqs.4 From eqs. 1  F wall = f From eqs.3  f =  s mg  F wall =  s mg We substitute F wall into eqs.4  d =  s mgLsin  /mgcos  =  s Ltan  = 0.51  3  tan(58  ) = 2.4 m (11-18)

19. Example (11-5) page 307 The biceps muscle bends the arm. Lengths a = 30 cm and x = 4 cm. If a mass M is held in your hand with the forearm horizontal and the upper arm vertical, determine the upward force F B the biceps exerts on the forearm bones  net = xF B – aMg = 0  F B = (a/x)Mg = (30/4)Mg = 7.5Mg The force F B exerted by the muscle is 7.5 times greater than the weight itself! (11-19) y xO CW CCW