CHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY As previously defined, an object is in equilibrium when it is at rest or moving with constant velocity, i.e., with no net force acting on it. The following are examples of objects in static equilibrium Conditions for static equilibrium Center of gravity (weight) Examples of static equilibrium Elasticity Stress and Strain Young s modulus Hooke s Law and the elastic limit Shear modulus
In each of those cases there is no translational motion and no rotational motion. Therefore, two conditions are necessary for an object to be in static equilibrium The net external force acting on the object is zero, i.e., i F i = 0, so there is no translational motion. The net external torque about any point is zero, i.e., i τ i = 0, so there is no rotational motion. Question 12.1: A see-saw consists of a board, 4.0 m in length with a mass of 20.0 kg. It supports two children whose masses are 28 kg and 40 kg. If the support (called the fulcrum) is under the center of the board, (a) What is the upward force exerted by the support on the board? (b) If the 28 kg child sits at one end of the board, where should the 40 kg child sit to balance the seesaw, i.e., to have the see-saw in static equilibrium? This is best illustrated by a problem
In fact, we could have taken torques about any point. 4 m The conditions for static and rotational equilibrium are i F yi = 0 and i τ i = 0. (a) The fulcrum must supply an upward force so the net force on the board is zero, i.e., F (20 kg)g (28 kg)g (40 kg)g = 0. F = (88 kg)g = 863.3 N. (b) For rotational equilibrium, define ccw torques as positive and sum the torques about the fulcrum, i.e., (28 kg)g (2 m) (40 kg)g d = 0. d = 56 kg m 40 kg = 1.40 m. So the 40 kg child should stand 1.40 m from the fulcrum. 4 m For example, taking torques about the left hand end we get the same answer (88 kg)g (2 m) (20 kg)g (2 m) (40 kg)g (d + 2 m) = 0 (40 kg) d = 56 kg m, 56 kg m i.e., d = = 1.40 m. 40 kg
Center of gravity (weight) We assumed that the weight force acts through the center of mass however, strictly speaking it should be the center of weight. Take some origin, O, and break the object into small elements of mass m i. The total torque about O is τ = i m i g i x i = Wx CG, where is the center of gravity (weight) defined as x CG x CG = i m i g i x i = i w i x i = i w i x i i m i g i i w i W. The center of mass is at the same point as the center of weight, i.e., x CM x CG, only if g is constant, then τ = ( i m i x i )g = Mgx CM = Wx CM. If the origin is taken at the center of gravity, then the object produces zero torque about that point. The center of gravity (weight) is then the point about which the gravitational force on the object produces zero torque no matter the orientation of the object.
DISCUSSION PROBLEM Why can you not touch your toes without falling over if you have you heels against a wall, as in (a) yet, ordinarily, you have no problem, as in (b)? If you don t believe me try it Question 12.2: A square plate is made by welding together four smaller square plates, each of side l. Each square is made from different materials so they have different weights, as shown in the figure. (a) Find the position of the center of gravity relative to the point (0,0). (b) If the plate is suspended from point P, what is the angle between the vertical and the left hand side of the plate?
P l l ( 0,0) 60 N 50 N (a) By definition, the center of gravity (weight), with respect to ( 0,0) is given by w x CG = i i x i W (60 N) l = 2 + (50 N) l 2 + (20 N) 3l 2 + (30 N) 3l 2 (60 N) + (50 N) + (20 N) + (30 N) and y CG = 20 N 30 N l l w i y i i W = 0.8125l, (60 N) 3l = 2 + (50 N) l 2 + (20 N) 3l 2 + (30 N) l 2 (60 N) + (50 N) + (20 N) + (30 N) (b) When hung from P, a vertical line through P passes through the CG. 0.8125l (0, 0) l φ CG P tan φ = 0.8125l, l i.e., φ = 39.1 ". = 1.000l. Therefore, the coordinates of the CG relative to ( 0,0) are (0.8125l, l).
1 m Question 12.3: A sign hangs in front of a store. The sign has a mass of 20 kg and it hangs at the end of a horizontal rod of length 2 m and mass 4 kg, which is hinged at the wall. The rod is supported by a wire attached to a point on the wall 1 m above the rod. (a) What is the tension in the wire? (b) What is the magnitude and direction of the force the rod exerts on the wall? 2 m 4 kg I physics 20 kg F y 1 m F F x Force of the wall on the rod T 2 m 4 kg (a) The wall must supply a force on the rod... how do we know that? However, we have no idea in what direction so choose it arbitrarily to begin with. Note, by Newton s 3rd Law the force the wall exerts on the rod is equal and opposite to the force the rod exerts on the wall. For static and rotational equilibrium: i F xi = 0, i F yi = 0 and = 0. i τ i Take torques about the hinge... a great idea since we don t know anything about F. Then T y (2 m) (4 kg)g (1 m) (20 kg)g (2 m) = 0 T y = Tx I physics 20 kg (44 kg m)g (2 m) T y y = 215.8 N. x
Fy 1 m F x F T 2 m 4 kg θ Tx I physics 20 kg Ty But T y (1 m) = tanθ = T x (2 m) = 1 2. T x = 2 T y, i.e., T x = 431.6 N. T = T x 2 + T y 2 = ( 431.6 N) 2 + (215.8 N) 2 = 482.5 N Note we found T even though we know nothing about F. Of course, the wire must be strong enough to withstand a force of 482.5 N. y x Fy 1 m F x F T 2 m 4 kg i F xi = 0, i F yi T x I physics 20 kg (b) For static and rotational equilibrium: = 0 and = 0. and F y So T y i τ i i F xi = F x + T x = F x 431.6 N = 0 F x = 431.6 N, i F yi = F y + T y (20 kg)g (4 kg)g = 0 F y = (24 kg)g (215.8 N) =19.6 N. F = (431.6ˆ i + 19.6ˆ j ) N φ F x φ = tan 1 F y = 2.6 " F x Therefore, the force of the rod on the wall is F = (431.6ˆ i +19.6ˆ j ) N. y x
1 m F T 2 m T x T y y x Fy F x 4 kg I physics 20 kg Note: we could find F another way... (a) by taking torques about the right hand end of the rod, since the system is in static equilibrium, i τ i = 0 about any point, i.e., (4 kg)g (1 m) F y (2 m) = 0. F y = (4 kg m)g (2 m) = 19.6 N. (b) By taking torques about the top end of the wire, i.e., F x (1 m) (4 kg)g (1 m) (20 kg)g (2 m) = 0. F x = (44 kg m)g (1 m) = 431.6 N. Question 12.5: A uniform ladder rests against a frictionless, vertical wall. If the coefficient of static friction between the bottom of the ladder and the floor is 0.3, what is the smallest angle at which the ladder can rest against the wall without slipping?
Let the ladder, length l, rest at an angle θ against the wall. Identify all of the force acting on the ladder. " F n θ " f s l " w " F 1 lsin θ Since the wall is frictionless, the force of the wall acting on the ladder is " F 1, which is to the wall, i.e., a normal force. The ground exerts two forces on the ladder; a l cosθ l cosθ normal force F " 2 2 n and a static frictional force f " s. The weight of the ladder is w ", which acts through the CG, i.e., the middle of the ladder. Taking torques about the foot of the ladder: i τ i = w l 2 cosθ + F 1(lsin θ) = 0... (3) F n θ f s l w F 1 lsin θ tanθ = w 2F 1. But from (1) and (2): F 1 = f s = µ s F n = µ s w. θ = tan 1 1 2µs = 59.0 #. ( ) For static and rotational equilibrium: and i F xi = 0, i F yi = 0 and # = 0. i τ i = f s F 1 = 0......... (1) F xi i i F yi = F n w = 0......... (2) l cosθ 2 l cosθ 2
Again, identify all the forces acting on the ladder. Summing the vertical and horizontal forces we get F n F 1 f s w and F y = F n w + f s... (1) F x = F 1......... (2). Question 12.6: In the previous problem there was no friction at the wall but there was static friction between the ground and the foot of the ladder. If, instead, there was friction at the wall but no friction at the ground, what difference would it make? Now eq (1) can be zero (if f s = w F n ). However, eq (2) can never be zero if the ladder is leaning against the wall as it consists of only one term ( F 1 ). Consequently, the ladder can never be in equilibrium, i.e., if there s no friction with the ground it will always slip We ignore the arrangement when the ladder is vertical
Edge of the table Question 12.7: A box X, of mass 10 kg, is placed on a box Y, of mass 5 kg, so that the center of box X lies directly over the left hand edge of box Y. If the two boxes are then placed over the edge of a table, what is the maximum possible overhang, h, without the boxes falling, if the sides of the boxes is 0.50 m? h L 2 x CG 10 kg 5 kg For balance the CG of the two boxes combined must lie directly over the edge of the table, i.e., the pivot point. Let the length of the side of the boxes be L, and find the CG with respect to the center line of the upper box (X), which is the same as the left hand edge of lower box (Y): x CG = L 2 m i gx i i m i g i = (10 kg)g 0 + (5 kg)g L 2 (15 kg)g = L 6. Therefore, total overhang is, h = L 6 + L 2 = 2L 3 = 0.33 m.
1.0 m 2.0 m Question 12.8: The figure shows a cart of mass 100 kg loaded with six identical packing cases, each of mass 50 kg. How is the total weight of the cart and packing cases distributed between the left hand and the right hand wheels?
Stress and strain (Young s modulus) At equilibrium 2.0 m 1.0 m 150 kg 100 kg 50 kg F L 100 kg F R F y = 0 and τ = 0. (F L + F R ) (150 kg)g (100 kg)g (100 kg)g (50 kg)g = 0, i.e., (F L + F R ) = 3924 N. Taking torques about the axis of the right hand wheel (150 kg)g (2 m) + (100 kg)g (1 m) +(100 kg)g (1 m) F L (2 m) = 0. F L = (500 kg m)g (2 m) = 2452.5 N so F R = 3924 N 2452.5 N = 1471.5 N. Objects deform when subjected to a force. In the case of the length of a metal bar or wire... The stress F A, and the strain ΔL L. Young s modulus Y = stress strain = Units: Force/area N m 2 (scalar) Thomas Young (1773-1829) ( F A ) ( ) ΔL L The example above is tensile stress. There is also compressive stress. Young s modulus is often the same in both cases, but there are exceptions (e.g., bone and concrete). L F F ΔL + L Area A
Question 12.9: A steel wire of length 1.50 m and diameter 1.00 mm is joined to an aluminum wire of identical dimensions to make a composite wire of length 3.00 m. (a) What is the length of the composite wire if it is used to support a mass of 5 kg? (b) What is the maximum load the composite wire could withstand? Assume the weights of the wires are negligible. Aluminum: Y Al = 70 109 N m 2. Tensile strength = 90 10 6 N m 2. Steel: Y steel = 200 109 N m 2. Tensile strength = 520 10 6 N m 2.
(a) The Young s modulus is Y = stress strain = 1.50m aluminum 1.50m d = 1.00mm 5kg steel ( F A ) ( ). ΔL L The stress in each wire is: F (5 kg)g A = π (0.5 10 3 m) 2 = 6.25 10 7 N m 2. But ΔL = ( L Y) ( F A). ΔL = (6.25 10 7 N m 2 ) L Y. Aluminum: ΔL 1 = (6.25 10 7 N m 2 (1.50 m) ) (70 10 9 N m 2 ) = 1.34 10 3 m 1.34 mm. Steel: ΔL 2 = (6.25 10 7 N m 2 (1.50 m) ) (200 10 9 N m 2 ) = 0.47 10 3 m 0.47 mm. (b) The tensile strength of aluminum is less than that of steel. Consequently, as the load increases, it will be the aluminum wire that fails. The maximum stress is F A = Mg A = 90 106 N m 2. M = (90 10 6 N m 2 ) A g = (90 10 6 N m 2 ) π(0.5 10 3 m) 2 9.81 m/s 2 = 7.21 kg. Therefore, the increase in length is 0.00181 m, so the new length is 3.00181 m.
Shear modulus F Area = A 60 mm 5 mm L θ F x 20 mm Δx F The shear strain is Δx L = tanθ The stress is F A Shear modulus M s = = θ (note which area) shear stress shear strain ( F A ) ( F ( ) = A ) Δx L tanθ. Question 12.10: A block of gelatin is 60 mm by 60 mm by 20 mm thick when unstressed. A force of 0.245 N is applied tangentially to the upper surface producing a 5 mm displacement relative to the lower surface. Find (a) the shearing stress, (b) the shearing strain, and (c) the shear modulus. Units: Force/area N m 2 (scalar).
60 mm 5 mm 20 mm θ (a) The stress is F A = 0.245 N (60 10 3 m) 2 = 68.1 N m 2 (Pascals - Pa). (b) The strain is (c) The shear modulus is tanθ = 5 mm 20 mm = 0.25. M S = stress 68.1 N m 2 = strain 0.25 = 272.4 N m 2.