2. Topic 2.2 Extended B – Applications of Newton’s second

3.  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

4. 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

5. 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

6. Topic 2.2 Extended B – Applications of Newton’s second

7.  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

8. 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."

9. 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?

10. 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.

11. 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.

12. 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.

13. 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

14. 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:

15. 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

16. 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.

17. 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

18. 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.

19. 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:

20. 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

21. 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.

22. 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

23. 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?

24. 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