Download presentation

1
**ENGR-1100 Introduction to Engineering Analysis**

Lecture 4

2
Lecture Outline Free body diagram. Equilibrium of a particle - two dimensional problems.

3
A Particle Large 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=0 Where SF is the sum of all forces acting on a particle.

4
Free Body Diagram A 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. Fcd mg

5
**The 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.

6
**What should be the body of “interest” of the following problem?**

or Fcd Fac Fb mg Fcd

7
**How 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. N2 N1 mg

8
**Procedure 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.

9
**Equilibrium of a Particle – Two Dimensions**

R= Rx + Ry = Rn + Rt = Rxi+ Ryj= Rnen+ Rtet =SFxi+ SFyj= SFnen+ SFtet =0 The equation can only be satisfied if: Rx = Rxi=SFxi=0 Ry = Ryj=SFyj=0 Rn = Rnen=SFnen=0 Rt = Rtet=SFtet=0 or

10
Example – P3-4 A 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.

11
**Solution 1) Free body diagram a) Choose the body**

b) Prepare the drawing c) Identify all forces mg FCB FAB a tan(a)=100/ a=140 d) Choose coordinate axes y x

12
**FCB = FAB sin(a)=2525.4*sin(140)**

1) Equilibrium equations mg FAB FCB y x a How many unknowns? How many equations? 2 2 SFx=0 FCB-FAB sin(a)=0 FCB = FAB sin(a)=2525.4*sin(140) FCB=610.9 N SFy=0 FAB cos(a)-250*9.8=0 FAB = 250*9.8/cos(a) = N

13
**Class Assignment: Exercise set 3-3**

please submit to TA at the end of the lecture A 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.

14
**Solution 1) Free body diagram a) Choose the body**

b) Prepare the drawing c) Identify all forces FB 300 y x d) Choose coordinate axes FA mg

15
**SFx=0 FA-FB sin(300)=0 SFy=0 FB cos(300)-500=0**

mg FA y x FB 300 1) Equilibrium equations How many unknowns? How many equations? SFx=0 FA-FB sin(300)=0 FA = FB sin(300)=577.4*sin(300) FA=288.7 lb SFy=0 FB cos(300)-500=0 FB = 500/cos(300) = lb

16
Example – P3-17 A 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. A B

17
**Solution SFy=0 T -75=0 T =75 lb 1) Free body diagram – body #B T y**

18
**1) Free body diagram – body #A**

T a b T y x 50 lb How many unknowns? How many equations? Two: a and b Two: SFx=0, SFy=0

19
**SFx=0 a=b T cos(a) - T cos(b) =0 SFy=0 T sin(a)+T sin(a) - 50=0**

50 lb SFx=0 a=b T cos(a) - T cos(b) =0 SFy=0 T sin(a)+T sin(a) - 50=0 sin(a) = 50/150 a = 19.50 y=10/2*tan(19.50)=1.77 ft

20
**Class Assignment: Exercise set 3-13**

please submit to TA at the end of the lecture Two 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 lb a=6.340 TBC=9.06 lb B A

Similar presentations

Presentation is loading. Please wait....

OK

5.3 Equations of Equilibrium

5.3 Equations of Equilibrium

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on 21st century skills for 21st Ppt on bank lending policy Ppt on word association test images Ppt on cse related topics to accounting Ppt on famous temples of india in hindi Holographic 3d display ppt on tv Download ppt on working of human eye Ppt on atm machine working Free download ppt on autonomous car Ppt on automobile related topics about work