Objective: Equilibrium Applications of Newton s Laws of Motion I

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1 Type: Single Date: Objective: Equilibrium Applications of Newton s Laws of Motion I Homework: Assignment (1-11) Read ( , 4.8, 4.11); Do PROB # s (46, 47, 52, 58) Ch. 4

2 AP Physics B Mr. Mirro Equilibrium, Statics and Newton s Laws I Date: WHAT IS THE E X A C T CONNECTION BETWEEN FORCE AND MOTION? Aristotle (300 B.C.) believed that the natural state of a body was at rest, and that a force was required to keep an object in motion along a horizontal plane. Galileo Galilei (1600 A.D.) maintained that it was just as natural for an object to be in horizontal motion with constant velocity as it is for it to be at rest! Upon this foundation, Sir Isaac Newton (1700 A.D.) built his great theory of motion. 1. Newton s First Law: Law of Inertia An object continues in a state of rest or in a state of motion at a constant speed along a straight line, unless compelled to change that state by a net force. Inertia is the tendency of a body to maintain its state of motion. 2. Newton s Second Law: Law of Acceleration When a net external force Σ F acts on an object of mass m, the acceleration a that results is directly proportional to the net force and indirectly proportional to the mass. That is, a = Σ F [a.k.a. f = ma] m Force M A-C-C-E-L-E-R-A-T-I-O-N 3. Newton s Third Law: Law of Action-Reaction Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. That is, F 1-on-2 = ( --) F 2-on-1 F 1-on-2 (--) F 2-on-1 M 1 M 2

3 When the forces exerted on a system cancel one another so that there is no change in velocity or no acceleration, the system is in equilibrium - there is a balance the study of which is called statics. An object is in EQUILIBRIUM under the action of concurrent forces provided it is NOT accelerating. The conditions for the equilibrium of a rigid body must include translational equilibrium and rotational equilibrium. The first condition for equilibrium is the requirement that the sum of the forces or the net force is equal to zero. Σ F = 0 The second condition for equilibrium is the requirement that the sum of the torques or the net torque is equal to zero. Σ τ = 0 Note: We will concentrate on the former as we discuss equilibrium and Newton s Laws of Motion. The concept of equilibrium arises directly from Newton s Second Law of Acceleration or F = ma. Σ F = m * a 0 Σ F x = m * a x = 0 Σ F y = m * a y = 0 These two equations are applicable independently and simultaneously. Strategy Isolate the object for discussion. Show the forces acting on the isolated object in a free-body diagram. Find the rectangular components of each force. Write the first condition for equilibrium in equation form. Solve for the required quantities.

4 AP Physics B Mr. Mirro Equilibrium, Statics and Newton s Laws I (cont.) Date: Consider the most basic example, A friend gives you a special gift, a box of mass 10 kg with a mystery surprise inside. It s a reward for your fine showing on the physics final examination. The box is resting on a smooth frictionless horizontal surface of a table. Use the conditions of equilibrium and Newton s Laws to analyze the following: a. Determine the weight of the box and the normal force acting on it. F x = 0 + F y = 0 +N + (-W) = 0 +N - mg = 0 +N = mg 10 kg (10 m/s 2 ) = 100 N N W b. Now your friend pushes down with a force of 40 N, again determine the normal force acting on it. F x = 0 + F y = 0 +N + (-W) + (- F) = 0 N - mg - 40 = 0 N = N 140 N N W 40 N c. If your friend pulls upward on the box with a force of 40 N, what now is the normal force on the box? F x = 0 + F y = 0 +N + (-W) + F = 0 N - mg + 40 = 0 N = N 60 N N W 40 N Now, although you knew how to do this problem intuitively in the past - this formal process of applying Newton s 2 nd Law can be applied now to the GENERAL case!

5 Now let s apply this process to more complicated example. Consider the net force exerted on a ring by three people. If the ring is in equilibrium there should be zero net force! 1. Free-body diagram +y Let: F 1 F 2 θ 1 θ 2 -x +x θ 1 = _60 _, θ 2 = _45 _ F 1 = 707 N F 2 = 500 N F 3 = 966 N Σ F =?? F 3 -y 2. Table of force components Force x-component y-component F 1 - F 1 cos θ cos 60 = -354 N + F 1 sin θ sin 60 = 612 N F 2 + F 2 cos θ cos 45 = N + F 2 sin θ sin 45 = 354 N F 3 0 -F 3 = -966 N 3. Apply the first condition for equilibrium: Σ F = 0 <i> Σ F x = 0 <ii> Σ F y = 0 F 1x + F 2x + F 3x = 0 F 1y + F 2y + F 3y = N N + 0 N = N N 966 N = 0 Are the sum of the forces nearly zero?? [YES or NO] Explain - E Q U I L I B R I U M 4. Depending on the question, we can set up simultaneous equations, solve for one quantity in terms of the others, simplify and find an answer.

6 AP Physics B Mr. Mirro Equilibrium, Statics and Newton s Laws I Date: Ex: A rope extends between two poles as shown below. If a 90 N youngster hangs off centered from the rope making angles of 10 and 5 respectively, find the tension in the two parts of the rope. [Schaums4.2] 10 5 T 1 T 2

7 Ex:

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