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Topic 2.2 Extended B – Applications of Newton’s second
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Newton’s 2 nd law is all about forces acting on masses. In this section you’ll be introduced to some of the basic forces of mechanics. If you drop a mass m very close to the surface of the earth it will fall, satisfying the 2 nd law: ∑F = ma x y x y W = ma W = mg W = -mgy |W|= W = mg W = mg weight Weight always points to the center of Earth. vector form of weight scalar form of weight W W W The fbd
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Topic 2.2 Extended B – Applications of Newton’s second Another force you are familiar with is the normal force N. This is the surface contact force that you feel when you push on a solid, such as a block of wood, a wall, or a floor. The normal force always acts perpendicular to the contact surface, and points away from it. W W Note that the weight always points to the center of Earth, but the normal force points perpendicular to the contact surface. N N Note that the normal force only exists if the mass is in contact with a surface. But the weight is always present. W The fbd N W N
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Topic 2.2 Extended B – Applications of Newton’s second Yet another force you are familiar with is the friction force f. If you attempt to slide an object to the right it will resist you with friction. The friction always acts in a direction opposite to the pull and parallel to the contact surface. Thus the two surface contact forces are friction and the normal. They are perpendicular to each other. W your pull to the right W the friction force f N W The fbd N f
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Topic 2.2 Extended B – Applications of Newton’s second
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A fourth force you are familiar with is the “tug of war” force (aka the “pulling” force.) A pull along the length of a wire or a rope or a pipe is called a tension T. W T the tension W N f Note that the tension appears only when the rope is tightened. Tension points in the direction of the line of the rope, away from the object being pulled. W N You cannot create a pushing force (compression) with a rope. W The fbd N f T
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Topic 2.2 Extended B – Applications of Newton’s second Suppose a 5-kg crate is to be dragged across a virtually frictionless ice surface by a rope inclined at 30°. The pull is 40-n. (a) What is the acceleration of the crate? W T N 30° W The fbd T 30° x y N ∑F x = ma x T cos 30° = 5 axax axax a x = 6.93 m/s 2 40 cos 30° T cos 30° T sin 30° We say that "T has been resolved."
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Topic 2.2 Extended B – Applications of Newton’s second Suppose a 5-kg crate is to be dragged across a virtually frictionless ice surface by a rope inclined at 30°. The pull is 40-n. (b) What is value of the normal force? W T N 30° W The fbd T 30° x y N ∑F y = ma y + - WN= 5 ayay + - WN= 5 0 mg N = T cos 30° T sin 30° + T sin 30° + 40 sin 30° - 20 5(10) N = - 20 30 n N = The weight of the crate is 50 n. Why isn't the normal force equal to the weight?
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Topic 2.2 Extended B – Applications of Newton’s second Suppose a 5-kg crate is suspended by a spring scale from the top of an elevator which is accelerating upward at 2 m/s 2. W T a The fbd y T a W What is the “apparent weight” of the crate (as read by the spring scale)? ∑F y = ma y T= 5 ayay - W T= 5 ayay - mg T= 5 ayay + mg T= 5 (2) + 5(10) T= 60 n The spring scale reads the tension. Since the “real weight” of the crate is 50 n, it appears to have “gained” 10 n. An observer in the elevator (a non- inertial frame) would not know of the elevator’s acceleration and would think that 60-n was the actual weight. Officially, accelerations are not drawn from the body. If included as a reference in the fbd, have the acceleration off to the side.
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Topic 2.2 Extended B – Applications of Newton’s second Many problems in physics involve more than one body. As an example, consider the system shown here: m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 When released from rest, the two masses will begin to accelerate. If the string doesn’t change length, their accelerations will be the same. a a If the pulley is light enough, we don’t have to take into account its presence in the system as a third accelerating body. Thus, we consider this system as a two- body system, with each body having the same acceleration. All the pulley does is redirect the tension.
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Topic 2.2 Extended B – Applications of Newton’s second Since there are two bodies, there are two free body diagrams. m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 m1m1 m2m2 a a W1W1 N1N1 T a W2W2 T f FBD, m 1 FBD, m 2 a Each body has a weight. Each body has a tension. Tension has the property that it is the same everywhere in a cord (if it is light). Therefore, only one symbol T is needed. Body 1 has a normal force, because it is in contact with a solid surface. Body 1 also has friction, acting in opposition to the rightward pull of the tension. And for book-keeping purposes we can put the accelerations in the FBDs. Note only one symbol is needed, since they are the same.
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Topic 2.2 Extended B – Applications of Newton’s second Each body will satisfy the 2 nd law: m1m1 m2m2 a a W1W1 N1N1 T a W2W2 T f FBD, m 1 FBD, m 2 a Body 1: Body 2: ∑ F x = m 1 a x ∑ F y = m 1 a y N 1 – W 1 = m 1 (0) T – f = m 1 a N 1 = W 1 T = m 1 a + f Suppose we want to find the values of T and a: ∑ F y = m 2 a y T – W 2 = m 2 (-a) T – m 2 g = -m 2 a …and of no use in this problem… expected T = m 2 g - m 2 a
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Topic 2.2 Extended B – Applications of Newton’s second m1m1 m2m2 a a W1W1 N1N1 T a W2W2 T f FBD, m 1 FBD, m 2 a Body 1: Body 2: T = m 1 a + f The final equations for each body were: T = m 2 g - m 2 a To find out what a is in terms of the other quantities, set the two equations equal (eliminating the T): m 1 a + f = T = m 2 g - m 2 a m 1 a + f = m 2 g - m 2 a m 1 a + m 2 a = m 2 g - f (m 1 + m 2 )a = m 2 g - f m 2 g – f m 1 + m 2 a = Note that we need values for m 1, m 2, and f in order to find a:
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Topic 2.2 Extended B – Applications of Newton’s second m1m1 m2m2 a a W1W1 N1N1 T a W2W2 T f FBD, m 1 FBD, m 2 a Body 1: Body 2: T = m 1 a + f The final equations for each body were: T = m 2 g - m 2 a Suppose m 1 is 2 kg and m 2 is 3 kg and there is no friction. What are a and T? m 2 g – f m 1 + m 2 a = You can use either equation for T: (3)(10) – 0 2 + 3 = = 6 m/s 2 T = m 1 a + f = (2)(6) + (0) = 12 n What is the maximum friction force you can have before the masses do not accelerate? (3)(10) – f 2 + 3 0 = f = 30 n
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Topic 2.2 Extended B – Applications of Newton’s second mgmg T3T3 FBD, knot FBD, m Sometimes problems will involve bodies that are not even moving, and might not be masses in the conventional sense: Suppose three cords support a crate of mass m as shown. knot A knot is at the intersection of the three cords. 30° 45° If we know the geometry of the system (the angles), we can find all of the tensions. Since the three ropes all do different things, their tensions will all be different. Not only can we make a free body diagram of the crate, but we can make one for the knot. T1T1 T2T2 T3T3 T2T2 T1T1 T3T3 30° 45° No book-keeping acceleration is needed, since nothing is moving.
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Topic 2.2 Extended B – Applications of Newton’s second mgmg T3T3 FBD, knot FBD, m From the free body diagram for the crate of mass m: T 3 - mg = 0 knot From the free body diagram for the knot: 30° 45° T1T1 T2T2 T3T3 T2T2 T1T1 T3T3 30° 45° T 3 = mg ∑F x = 0 ∑F y = 0 T 2 cos 45° - T 1 cos 30° = 0 T 2 sin 45° + T 1 sin 30° - T 3 = 0 0.707T 2 + 0.5T 1 = mg T 2 = 1.225T 1 0.707(1.225T 1 ) + 0.5T 1 = mg 0.707T 2 - 0.866T 1 = 0 T 1 = mg/1.366 T 2 = 1.225(mg/1.366) T 2 = 0.897mg
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Topic 2.2 Extended B – Applications of Newton’s second mgmg T3T3 FBD, knot FBD, m Our results: knot If the weight of the crate is 20 n, what are the three tensions? 30° 45° T1T1 T2T2 T3T3 T2T2 T1T1 T3T3 30° 45° T 3 = mg T 1 = mg/1.366 T 2 = 0.897mg W = 20 n = mg T 3 = 20 n T 1 = 20/1.366 = 14.64 n T 2 = 0.897(20) = 17.94 n Why don’t T 1 and T 2 add up to T 3 ? Is it possible to have a geometry such that T 1 and T 2 each support exactly half the crate’s weight.
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Topic 2.2 Extended B – Applications of Newton’s second y x m1gm1g T FBD, m 2 FBD, m 1 Sometimes you have to do a little analysis to decide which way the masses will accelerate. If you choose an arbitrary direction, consistent with all bodies, you can still solve the problem. N T m2gm2g 30° m2m2 m1m1 Either of these choices is consistent. a a a a Suppose W 1 = 10 n, W 2 = 30 n, and that there is no friction. I will assume the acceleration of the system is as shown. The FBDs are therefore: a a y x This cs requires you to break down 3 vectors: a, T, and N. This cs requires you to break down only 1 vector, m 2 g Here is our reference angle:
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Topic 2.2 Extended B – Applications of Newton’s second a a y x m1gm1g T FBD, m 2 FBD, m 1 For m 1 : N T m2gm2g 30° m2m2 m1m1 a a T – m 1 g = - m 1 a W 1 = m 1 g 10 = m 1 (10) m 1 = 1 kg T – 10 = - 1a T = 10 + - 1a For m 2 : W 2 = m 2 g 30 = m 2 (10) m 2 = 3 kg m 2 g sin 30° - T = -m 2 a x-direction 30 sin 30° - T = - 3a T = 15 + 3a 10 + - 1a = 15 + 3a 4a = - 5 a = - 1.25 m/s 2 T = 10 + - 1a T = 10 + - 1( - 1.25) T = 11.25 n
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Topic 2.2 Extended B – Applications of Newton’s second a a y x m1gm1g T FBD, m 2 FBD, m 1 Interpreting the results: N T m2gm2g 30° m2m2 m1m1 a a Since our acceleration turned out to be negative, we know that we picked the wrong direction. Thus, m 1 really goes upward, and m 2 goes down the incline. The numeric values we got for a and T during the analysis are still valid.
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Topic 2.2 Extended B – Applications of Newton’s second a a y x m1gm1g T FBD, m 2 FBD, m 1 An alternate solution: N T m2gm2g 30° m2m2 m1m1 a a Since both masses are connected by the cord and move as one, we can treat them as one mass (of 1 + 3 kg): The only forces we need are the ones that are in the direction of motion. In the case of m 1 that force is m 1 g = 10 n. m2m2 m1m1 10 15 In the case of m 2 that force is m 2 g sin 30° = 15 n. the winner 15 - 10 = (1 + 3)a a a = 1.25 m/s 2
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Topic 2.2 Extended B – Applications of Newton’s second Two masses M and m are connected by a very light cord over a very light pulley, as shown. Assume M > m. m M Suppose we want to know what the acceleration of the blocks is, once released from rest. mgmg T FBD, m a MgMg T FBD, M a T – mg = ma For mFor M T – Mg = - Ma T = ma + mg T = Mg - Ma ma + mg = Mg - Ma ma + Ma = Mg - mg (m + M)a = (M – m)g a = g M – m M + m Question: If m = M, what does our equation predict for a?
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Topic 2.2 Extended B – Applications of Newton’s second Finally, we can tie together Newton's second law with the kinematic equations by the acceleration a: Suppose a 750-kg car that is traveling at 20 m/s slams on the brakes and comes to a halt after traveling 125 m. What was the friction force between the tires and the pavement during the braking? Since time is not given or required, use v 2 = v o 2 + 2ad 0 2 = 20 2 + 2a(125) a = -1.6 m/s 2 kinematics F = ma F = (750)( - 1.6) F = 1200 n dynamics
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