When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero. 0 0 0 0 k M j M i M M k R j R i R F R z y x z y x
Forces and moments acting on a rigid body could be external forces/moments or internal forces/moments. Forces acting from one body to another by direct physical contact or from the Earth are examples of external forces. Fluid pressure acting to the wall of a water tank or a force exerted by the tire of a truck to the road is all external forces.
The weight of a body is also an external force. Internal forces, on the other hand, keep the particles which constitute the body intact. Since internal forces occur in pairs that are equal in magnitude opposite in direction, they are not considered in the equilibrium of rigid bodies. The first step in the analysis of the equilibrium of rigid bodies must be to draw the free body diagram of the body in question.
Common Support / Connection Element Types in Two Dimensional Analysis In rigid bodies subjected to two dimensional force systems, the forces exerted from supports and connection elements are shown in the free body diagram as follows: It should be kept in mind that reaction will occur along the direction in which the motion of the body is restricted.
Equations of Equilibrium in Two Dimensional Case If all the forces acting on the rigid body are planar and all the couples are perpendicular to the plane of the body, equations of equilibrium become two dimensional. 0 0 0 0 k M M F R F R j R i R F R z y y x x y x or in scalar form, 0 0 0 O y x M F F At most three unknowns can be determined.
Alternative Equations of Equilibrium In two dimensional problems, in alternative to the above set of equations, two more sets of equations can be employed in the solution of problems. Points A, B and C in the latter set cannot lie along the same line, if they do, trivial equations will be obtained. 0 0 0 0 0 0 C B A B A x M M M M M F
Two-Force Member Members which are subjected to only two forces are named as two force members. Forces acting on these members are equal in magnitude, opposite in direction and are directed along the line joining the two points where the forces are applied. A B P A x A y B y B x F A F B F A =F B Weight is neglected. If weight is considered, the member will not be a two force member! P P Examples of two force members P P P Hydraulic cylinder P
Three-Force Member In rigid bodies acted on by only three forces, the lines of action of the forces must be concurrent; otherwise the body will rotate about the intersection point of the two forces due to the third force which is not concurrent. If the forces acting on the body are parallel, then the point of concurrency is assumed to be in infinity. F A P F B A P B
Free Body Diagram (FBD) The procedure for drawing a free body diagram which isolates a body or system consists of the following steps: 1) If there exists, identify the two force members in the problem. 2) Decide which system to isolate. 3) Isolate the chosen system by drawing a diagram which represents its complete external boundary. 4) If not given with the problem, select a coordinate system which appropriately suits with the given forces and/or dimensions. 5) Identify all forces which act on the isolated system applied by removing the contacting or attracting bodies, and represent them in their proper positions on the diagram. 6) Write the equations of equilibrium and solve for the unknowns.
A y A x M O O y O x
B x A y A x
N x N y R x R y
B y B x A x A y M A A x
T W=mg F f W=mg N N
T A x A x N L A y A y mg m O g
mg T A y B y A x B x F f N
B x B y T T A y A x A x A y mg L
1. A 54 kg crate rests on the 27 kg pickup tailgate. Calculate the tension T in each of the two restraining cables, one of which is shown. The centers of gravity are at G 1 and G 2. The crate is located midway between the two cables.
2. The pin A, which connects the 200-kg steel beam with center of gravity at G to the vertical column, is welded both to the beam and to the column. To test the weld, the 80-kg man loads the beam by exerting a 300 N force on the rope which passes through a hole in the beam as shown. Calculate the torque (couple) M supported by the pin.
3. A portion of the shifter mechanism for a manual car transmission is shown in the figure. For the 8 N force exerted on the shift knob, determine the corresponding force P exerted by the shift link BC on the transmission (not shown). Neglect friction in the ball and socket joint at O, in the joint at B and in the slip tube near support D. Note that a soft rubber bushing at D allows the slip tube to self-align with link BC.
4. A large symmetrical drum for drying sand is operated by the geared motor drive shown. If the mass of the sand is 750 kg and an average gear-tooth force of 2.6 kn is supplied by the motor pinion A to the drum gear normal to the contacting surfaces at B, calculate the average offset x of the center of mass G of the sand from the vertical centerline. Neglect all friction in the supporting rollers.
5. It is desired that a person be able to begin closing the van hatch from the open position shown with a 40-N vertical force P. As a design exercise, determine the necessary force in each of the two hydraulic struts AB. The center of gravity of the 40-kg door is 37.5 mm directly below point A. Treat the problem as twodimensional.
C 375 mm D 100 mm G A 75 mm B 75 mm 450 mm 30 o 400 N 6. The bracket and pulley assembly has a mass of 40 kg with combined center of gravity at G. Calculate the magnitude of the force supported by the pin at C and roller at A when a tension of 400 N is applied in the vertical plane of the cable.
300 mm 30 o 100 mm 300 mm k=50 N/m 7. The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in place by the spring which has an unstretched length of 200 mm. The cocking lever rests against a smooth peg at B. Determine the magnitude of the support force at A and the normal force on the peg at B when the lever is in the position shown.
8. Plate AB contains a smooth parabolic slot. Fixed pins B and C are located at the positions shown in the figure. The equation of the parabolic slot is given as y = x 2 /160, where x and y are in mm. If it is known that the force input P = 4 N, determine the forces applied to the plate by the pins B and C and also the force output Q. Q 120 mm P A y B C 46 mm 20 mm x 140 mm 60 mm 40 mm
b Q 120 mm P=4 kn B F F M 60B A 0. 6C 0. 219Q y x y 0. 8C 0. 975Q y 0 C 0 0 7. 13Q 4 90 y B y 140 mm 60 mm 40 mm B 0 P B b C a a C P C sina Q sin b 0 y C cos a Q cos b 0 Tangent to the parabolic slot 46 mm 20 mm x 2 x y 160 dy 2x y dx 160 y x60 120 46 20 tan b 140 60 40 22. 5 B 60 Q sin b46 40tan b y b 12. 68 22. 5 o 0 2 60 22. 5 mm 160 tana 3 4 C 4. 494 N Q 5. 95 N By 2. 207 N