Dynamics Equilibrium

Equilibrium Word itself contains an obvious hint as to its meaning Equilibrium describes a condition when a dynamic system experiences no net change. When systems reach equilibrium they are said to be in balance. There are many forms of equilibrium In chemistry, when reactions reach equilibrium a balance is reached between the amount of reactants and products. As a result, the reaction appeared to cease. There was no further net change either forward or backward. When you ride in an airplane or dive under water you experience an unbalanced pressure between the air on either side of your ear drum. By moving your jaw, and opening the eustachian tubes in your ears, you allow the movement of air until equilibrium is reached and pressure is again balanced. In physics there are different types of equilibrium. However, in this particular course equilibrium will have a very specific meaning. Equilibrium will refer specifically to equal and opposite forces or torques. Torques will be taught later in the course, so for now equilibrium will mean equal and opposite force.

Equilibrium For now, equilibrium will refer specifically to equal and opposite forces. Forces are vector quantities. When forces are in equilibrium all the forces pointing in one direction must equal all the forces pointing in the opposite direction. Examples: What is the magnitude of the missing force, F ? When forces are in equilibrium, the net force (sum of force vectors) is zero. Fnet = ΣF = ma = 0 If the net force is equal to zero, then acceleration is also equal to zero. If acceleration is equal to zero, then object is either stationary of moving at constant velocity. This type of motion is governed by Newton’s 1st law of motion (Inertia). 10 N F m 6 N 10 N F m = 4 N = 10 N

Solving Force Problems In kinematic problems you create a variable list and then chose from a few standard equations to solve for missing quantities. However, force problems are very different. Each problem will seem unique and will appear to require a custom solution. One option is to memorize all the millions and millions of possible force problems. This method leads to low grades and failing scores on AP Exams and college exams. (Some of you will try it anyway) Those that plan to succeed in this course, on the AP Exam, and in college will instead develop a method that solves any and all force problems. (Do this instead) Here is a suggested approach, until you device a method that suits you best. Assess the problem: What direction are we concerned with? What type of motion is the object experiencing? How does this impact the net force? Diagram: Create a formal free body diagram as requested and/or a diagram containing only the relevant force vectors and components. Sum forces: Write an equation for the net force (sum of force vectors). Solve: Substitute known equations and values to solve the problem.

Example 1 A 10 kg mass is pulled along a rough surface at constant velocity by a rope exerting 20 N of force. Determine the magnitude of the friction force. 10 kg Assess: What direction is this problem concerned with? “along a rough surface” x-direction What type of motion is the object experiencing? “constant velocity” How is this relevant? The object is in dynamic equilibrium in the x-direction. This means that the acceleration and net force (sum of force vectors) in the x-direction is zero, ΣF = ma = 0 . For this to occur the horizontal forces are balanced and must be equal and opposite. At this point you should be able answer the question without following the remaining steps. However, not all problems will be this simple. Let’s use this easy problem to demonstrate the remaining steps.

Example 1 A 10 kg mass is pulled along a rough surface at constant velocity by a rope exerting 20 N of force. Determine the magnitude of the friction force. 10 kg Diagram: Some problems will require a formal free body diagram and other will not. Free body diagrams can NEVER include vector components or a vector representing the net force. If you prefer a alternate diagram you are allowed to draw one. An alternate diagram is considered to be scratch work on exams, and is always optional and secondary to the free body diagram. An alternate diagram can contain vector components and include the sign on each force vector. It can also be limited to only the vectors acting in the relevant direction. Free body diagram Alternate diagram −f +T f T N Fg This diagram is not necessary in this problem, but is shown as an example of a diagram containing only the vectors matching the relevant x-direction motion.

Example 1 A 10 kg mass is pulled along a rough surface at constant velocity by a rope exerting 20 N of force. Determine the magnitude of the friction force. 10 kg Assess: Moving in the x-direction, equilibrium, ΣF = ma = 0 Diagram: Sum forces: Using this info, write the key sum of vector forces equation. ΣF = T − f Note: The relevant (horizontal vectors) with correct signs were simply transferred from the alternate diagram into the equation. Solve: Substitute known equations (ΣF = 0) and values (T = 20 N) into the above key sum of vector forces equation, and then solve. 0 = 20 − f f = 20 N f T N Fg −f +T

Example 2 Determine the rope tension in each scenario shown. Assess Diagram Sum Forces Solve 10 kg 10 kg y-direction, stationary, equilibrium, balanced force, ΣF = ma = 0 y-direction, stationary, equilibrium, balanced force, ΣF = ma = 0 10 kg T Fg 10 kg T Fg

Example 3 A 10 kg mass hanging from two ropes. Assess: Stationary in both the x- and y- directions Equilibrium: ΣFx = 0 and ΣFy = 0 Diagram: Free body diagram Alternate diagram Sum forces: You may need to sum forces one or both directions. Solve: The x-direction is a bust (0 = 0), while the y-direction solves. Fg T −Fg +Ty Tx

Example 4 A 10 kg mass is suspended as shown. 2 60o 30o 10kg Assess: Stationary in both the x- and y- directions Equilibrium: ΣFx = 0 and ΣFy = 0 Diagram: Free body diagram Alternate diagram NOTE: Tension is NOT related to string/rope length. In this problem the shorter string has the greatest tension. The mass is hanging closer to the point where string 2 attaches to the ceiling. This creates more tension is this string. T2 Fg T1 −Fg +T2y +T2x +T1y −T1x 60o 30o

Example 4 A 10 kg mass is suspended as shown. −Fg +T2y +T2x +T1y −T1x 60o 30o Assess: Stationary in the x- and y- directions ΣFx = 0 and ΣFy = 0 Diagram: Alternate diagram is more helpful. Sum forces: This time both x and y are needed. Solve: Two equations with two missing variables. What should we do next? By now you have completed both Algebra 1 and Algebra 2. You are at least enrolled in pre-calculus (Algebra 2 again). This is no longer a physics problem. It is now both a math problem and your problem. Enjoy.