## Presentation on theme: "Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad"— Presentation transcript:

ENGINEERS MECHANICS STATICS CHAPTER 5 Lecture Notes: Professor A. Salam Al-Ammri Suhad Ibraheem Mohammed Al-Khwarizmi College of Engineering University of Baghdad Analysis of Structures Trusses Machines 2/17/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

Contents Introduction Machines Sample Problem 5.6.1 Sample Problem 5.6.2 Sample Problem Sample Problem 5.6.4 Sample Problem 5.6.5 2/17/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

Machines Machines are structures contain moving parts whose components are capable of relative motion when not attached to a support, designed to transmit and modify forces. Their main purpose is to transform input forces into output forces. 2/17/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

Sample Problem 5.6.1 To determine the center of gravity of a non-uniform bar that weights 24 N, an engineer suspends the bar from a wire at point A and supports it at point B on a knife edge that rests on a pan of a balance. A weight of 16 N is required to balance the bar in a horizontal position. Determine the horizontal distance d from point A to the center of gravity of the bar. draw a free-body diagram of the scale draw a free-body diagram of the bar. 2/17/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

Sample Problem 5.6.2 A tackle on the football team weighs 350 lb. In coming off his stance, he places his entire weight on the ball of his foot. In effect, his foot acts as a lever. The weight of the tackle is transmitted to his foot through the tibia. In tern the Achilles tendon pulls on the heal consider the static effects only, determine the force that the tendon must withstand. 2/17/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

Sample Problem 5.6.3 What forces are exerted on the bolt at E in Fig. as a result of the 150-Nforces on the pliers? α= tan -1(30/70) = 23.2° ΣFx= 0 , Dx+ Rcos α= 0 ΣF y= 0, D y − R sin α+ 150 = 0 ΣMB= (30)D y −(100)(150) = 0 D x= −1517 N , D y= 500N R= 1650N ΣMC= −(30)E−(30)D x= 0 E = −D x= 1517 N 2/17/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad