Presentation on theme: "ENGR-1100 Introduction to Engineering Analysis"— Presentation transcript:
1ENGR-1100 Introduction to Engineering Analysis Lecture 4
2Lecture OutlineFree body diagram.Equilibrium of a particle - two dimensional problems.
3A ParticleLarge bodies or small bodies can be referred to as particles when the size and shape of the body have no effect on the response of the body to a system of forces.The mass of the body can be assumed to be concentrated at a point and the shape can be neglected.A necessary and sufficient condition for equilibrium:R=SF=0Where SF is the sum of all forces acting on a particle.
4Free Body DiagramA carefully prepared drawing that shows a “body of interest” separated from all interacting bodies.The forces exerted by all other bodies must be determined and shown on the diagram.Fcdmg
5The procedure for drawing free-body diagram Essential Steps Make a decision regarding what body is to be isolated and analyzed.Prepare a sketch of the external boundary of the selected body.Present all forces, known and unknown, that are applied by other bodies with vectors in the correct position.
6What should be the body of “interest” of the following problem? orFcdFacFbmgFcd
7How to deal with unknown forces? If a force has a known line of action but unknown magnitude and direction, the direction of the force can be assumed.If both magnitude and direction of the force are unknown it might be convenient to show the two rectangular components of the force.N2N1mg
8Procedure to construct a complete and correct free-body diagram: 1) Decide which body or combination of bodies is to be shown on the free-body diagram.2) Prepare a drawing of the outline of the free-body.3) Carefully trace around the boundary of the free body and identify all the forces exerted by contacting or attracting bodies that were removed during the isolation process.4) Choose the set of coordinate axes to be used in solving the problem and indicate their direction on the free-body diagram.
9Equilibrium of a Particle – Two Dimensions R= Rx + Ry = Rn + Rt= Rxi+ Ryj= Rnen+ Rtet=SFxi+ SFyj= SFnen+ SFtet =0The equation can only be satisfied if:Rx = Rxi=SFxi=0Ry = Ryj=SFyj=0Rn = Rnen=SFnen=0Rt = Rtet=SFtet=0or
10Example – P3-4A homogeneous cylinder with mass of 250 kg is supported against a smooth surface by a cable as shown in Fig Determine the forces exerted on the cylinder by the cable and by the smooth surface at contact point C.
11Solution 1) Free body diagram a) Choose the body b) Prepare the drawingc) Identify all forcesmgFCBFABatan(a)=100/ a=140d) Choose coordinate axesyx
12FCB = FAB sin(a)=2525.4*sin(140) 1) Equilibrium equationsmgFABFCByxaHow many unknowns?How many equations?22SFx=0FCB-FAB sin(a)=0FCB = FAB sin(a)=2525.4*sin(140)FCB=610.9 NSFy=0FAB cos(a)-250*9.8=0FAB = 250*9.8/cos(a) = N
13Class Assignment: Exercise set 3-3 please submit to TA at the end of the lectureA homogeneous cylinder weighing 500 lb rests against two smooth planes that form a trough as shown in Fig. P3-3.Determine the forces exerted on the cylinder by the plane at contact A and B.
14Solution 1) Free body diagram a) Choose the body b) Prepare the drawingc) Identify all forcesFB300yxd) Choose coordinate axesFAmg
16Example – P3-17A continuous cable is used to support blocks A and B as shown in Fig. P Block A is supported by a small wheel that is free to roll on the cable. Determine the displacement y of block A for equilibrium if the weight of block A and B are 50 lb and 75 lb, respectively.AB
17Solution SFy=0 T -75=0 T =75 lb 1) Free body diagram – body #B T y
181) Free body diagram – body #A TabTyx50 lbHow many unknowns?How many equations?Two: a and bTwo: SFx=0, SFy=0
19SFx=0 a=b T cos(a) - T cos(b) =0 SFy=0 T sin(a)+T sin(a) - 50=0 50 lbSFx=0a=bT cos(a) - T cos(b) =0SFy=0T sin(a)+T sin(a) - 50=0sin(a) = 50/150a = 19.50y=10/2*tan(19.50)=1.77 ft
20Class Assignment: Exercise set 3-13 please submit to TA at the end of the lectureTwo flower pots are supported with cables as shown in Fig. P If pot A weighs 10 lb and pot B weighs 8 lb, determine the tension of the cables and the slope of cable BC.Answer:TAB=TCD=12.73 lba=6.340TBC=9.06 lbBA