this is a sample powerpoint presentation on the concept of EQUILIBRIUM...

1. THE CONCEPT OF EQUILIBRIUM

2. According to Rene Descartes, the human body is a mechanical system designed by the hands of God. HUMAN LEVER

3. Borelli, a student of Galileo, considered the human body as a system of levers.

4. In our body, the bones act as the rigid bar and the joints the fulcrum. It is through these so-called human levers that we are able to move, no matter how slight our movement is, to talk, walk, and even eat.

5. EQUILIBRIUM Is the condition of force where it is acted but simply cancelled out. These forces may be even large enough to cause permanent deformation.

6. STATICS It is concerned with the calculation of the forces acting on and within structures that are in equilibrium.

7. FORCE SYSTEMS Concurrent and Nonconcurrent Concurrent system occur when the lines of actions of the forces acting on a body intersect at a common point. Nonconcurrent system occurs when the forces are acting at different points.

8. Nonconcurrent forces may also me parallel and non parallel. The lines of action of parallel forces do not intersect.

9. PARTICLE AND RIGID BODY Distinction has to made whether the object being acted upon by a system of forces is a particle or a rigid body. A PARTICLE may be considered as a point of mass A RIGID BODY is an extended body in space that does not change its size and shape

10. TRANSITIONAL EQUILIBRIUM Equilibrium is a condition where there is no change in the state of motion of a body. An object in equilibrium may be at rest. It can also be in motion, provided that it moves with constant speed in the same direction.

11. An object that has no net force acting on it is said to be in TRANSLATIONAL EQUILIBRIUM. This condition of translational equilibrium is usually referred to as the first condition for equilibrium and may be expressed in the form: ∑ F = 0

12. SAMPLE PROBLEM You hang your picture frame by means of vertical string. Two strings in turn support this string. Each string makes 30° with an overhead horizontal beam. Find the tension in the strings. T 1 T 2 T 3 30° 30°

13. SOLUTION We are given w = 55N. The forces acting on the frame are shown. The tension T 3 in the rope pulls on the frame upward, while the weight of the frame acts downward. The frame will be in equilibrium when ∑ F = 0. Free body part of the frame T 3 - 55 N = 0 T 3 = 55 N 55N T 3

14. To determine T 1 and T 2 consider the point where the three ropes meet as a particle in equilibrium. Applying the first condition for equilibrium, we have: ∑ F x = 0: T 1 cos 30°- T 2 cos 30° = 0 T 1 cos = T 2 ∑ F y = 0: T 1 sin 30°+ T 2 sin 30° - 55N = 0 Since T 1 = T 2 we may replace T 2 by T 1 and solve the above equation for T 1 . ∑ F y = 0: T 1 (.5) + T 1 (.5) – 55 N = 0 T 1 = 55 N T 2 = 55 N T 1 T 2 30° 30° w

15. TORQUE In physics, a TORQUE (τ) is a vector that measures the tendency of a force to rotate an object about some axis The magnitude of a torque is defined as force times its lever arm . Just as a force is a push or a pull, a torque can be thought of as a twist. The SI unit for torque is newton meters (N m). In U.S. customary units, it is measured in foot pounds (ft·lbf) (also known as 'pounds feet'). The symbol for torque is τ , the Greek letter tau .

16. EXPLANATION The force applied to a lever, multiplied by its distance from the lever's fulcrum, is the torque. For example, a force of three newtons applied two meters from the fulcrum exerts the same torque as one newton applied six meters from the fulcrum. This assumes the force is in a direction at right angles to the straight lever. The direction of the torque can be determined by using the right hand rule: Using your right hand, curl your fingers in the direction of rotation, and stick your thumb out so it is aligned with the axis of rotation. Your thumb points in the direction of the torque vector.

18. PROBLEM For a position a, the lever arm is already given as 0.25m. The force will produce a clockwise rotation respect to O and therefore the torque is negative. t = - (60 N)(.25 m) = -15 Nm SOLUTION The resultant force = 7.0 – 7.0 N = 0

19. CENTER OF GRAVITY The center of mass or center of gravity is a useful concept when dealing with equilibrium problems. Each of the particles making up a body hast its own weight or mass. The mass or weight of the object is the sum of the masses or weights of these particles. The center of gravity of a body is the point where its entire weight maybe assumed concentrated.

20. CENTER of GRAVITY of a GROUP of BODIES The center of gravity of a group of bodies whose centers of gravity are known from a fixed point is defined as the sum of the products of weight of individual body and it center of gravity divided by the total weight. In symbols: X W 1 X 1 + W 2 X 2 …. W 1 + W 2 + ….. =

21. PROBLEM Find the center of gravity of a flat piece of wood shaped in the form of letter L. All dimensions are in meters. The weight per square meter of the wood is 2.4 N/m 2 . y 0 (X 1 , Y 1 ) 3.0 4.0 3.0 (X 2 , Y 2 ) x

22. SOLUTION The center of gravity of the wood will specified by two coordinates: x and y. We divide the plate into regularly shaped parts. Let A 1 be the area of the bigger rectangle and A 2 the area of the smaller rectangle. The center of gravity of A 1 labeled as (x 1 ,y 1 ) is the (1.5 m, 3.0 m). The center of gravity of A 2 labeled as (x2, y2) is (4.5 m, 1.0 m). A 1 = (3 m) (6 m) = 18 m2 W 1 = (2.4 kg/m2) (18 m2) = 43.2 N A 2 = (3 m)(6 m) = 18 m2 W 2 = (2.4 kg/m2) (6.m2) = 14.4 N X = (43.2 N) (1.5 m) + (14.4 N)(4.5 m) 43.2 N + 14.4 N = 2.3 m y = (43.2 N) (3 m) + (14.4 N)(1 m) 43.2 N + 14.4 N 2.5 m =

23. ROTATIONAL EQUILIBRIUM A necessary condition for a body to be in rotational equilibrium is that the sum of the torques with their proper signs about point must be zero. ∑ t = 0 The condition is known as the second condition for equilibrium .

24. SAMPLE PROBLEM A 120 N child and a 200 N child sit at the opposite ends of 4.00 m uniform seesaw pivoted at its center. Where should a 140 N child sit to balance the seesaw? FREE BODY DIAGRAM OF THE SEESAW 2.0 m 120 N 140 N R W 200 N

25. SOLUTION Let w represent the weight of the seesaw. R is the upward reaction at the pivotal point. The third child must sit on the same side of the seesaw as the 120 N child. Let x be the distance of the third child from the center of the seesaw. Applying the second condition for equilibrium and talking summation of torques about the center of the seesaw will eliminate the rotational effect of R and w .

26. + (120 N)(2.00 m) + (140)(x) – (200 N)(2.00 m) = 0 Solving for x , x = 1.14 m from the center on the side of the 120 N child.

27. STABILITY Three types of equilibrium: UNSTABLE – the great example of this is a cone that is balance on its apex but when disturbed slightly, it will fall over STABLE – the condition of an object to return it is original position when slightly disturbed. NEUTRAL – the condition where an object is lying on its side and displace but manages to remains its equilibrium about its new position