Chapter 4. Dynamics: Newton s Laws of Motion

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Arlene Benson
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1 Chapter 4 Dynamics: Newton s Laws of Motion

2 Types of Forces: An Overview Examples of Nonfundamental Forces -- All of these are derived from the electroweak force: normal or support forces friction tension in a rope

When an object is in contact with a surface there is a force acting on that object. The component of this force that is parallel to the surface is called the frictional force.

3 When an object is in contact with a surface there is a force acting on that object. The component of this force that is parallel to the surface is called the frictional force. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. cold welds

4 When the two surfaces are not sliding across one another the friction is called static friction. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

5 The magnitude of the static frictional force can have any value from zero up to a maximum value. f f MAX s s MAX s f = µ s F N Not a vector equation! f S is parallel to the surface, F N is perpendicular to the surface. 0 < µ <1 is called the coefficient of static friction. s

6 Note that the magnitude of the frictional force does not depend on the contact area of the surfaces. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

7 Static friction opposes the impending relative motion between two objects. Kinetic friction opposes the relative sliding motion that actually does occur. f = µ k k F N 0 <1 < s µ is called the coefficient of kinetic friction. k

8 The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. Usually, µ s > µ k

9 Example. A sled and a rider are moving at a speed of 4.0 m/s along a horizontal stretch of snow. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is and the mass of the sled and rider is 40 kg. Find the kinetic frictional force and the displacement, x, of the sled. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

10 1. Use Newton s 2nd law in x and y directions. ΣF x = -f k = ma x --> a x = -f k /m = -µ k F N /m (since f k = µ k F N ) ΣF y = F N - W = F N - mg = ma y = 0 --> F N = mg f k = µ k F N = µ k mg = (0.050)(40)(9.8) = 20 N

11 a x = -µ k F N /m = -µ k mg/m = -µ k g = -(0.050)(9.8) = m/s 2 2. Solve for x using a x and kinematic equations. x v 0x v x a x t? 4.0 m/s 0 m/s m/s 2 v x 2 = v 0x 2 + 2a x x --> x = (v x 2 - v 0x2 )/(2a x ) = ( )/(2(-0.49)) = 16 m independent of mass of sled+rider

Example. Block 1 (mass m 1 = 8.00 kg) is moving on a 30.0 o incline with a coefficient of kinetic friction between the block and incline of 0.300.

12 Example. Block 1 (mass m 1 = 8.00 kg) is moving on a 30.0 o incline with a coefficient of kinetic friction between the block and incline of This block is connected to block 2 (mass m 2 = 22.0 kg) by a massless cord that passes over a massless and frictionless pulley. Find the acceleration of each block and the tension in the cord. f k

13 y! a x y f k x f k! a m 1 = 8.00 kg m 2 = 22.0 kg µ k = Find a, T and T

14 Block 1: y! a x Σ F x = -f k - W 1 sin 30.0 o + T = m 1 a Σ F y = F N - W 1 cos 30.0 o = 0 è F N = W 1 cos 30.0 o y f k Block 2: Σ F y = T - W 2 = m 2 (-a) We also know: x! a T = T since the pulley and cord are massless f k = µ k F N = µ k m 1 g cos 30.0 o = (0.300)(8.00)(9.80)(0.866) = 20.4 N

15 Equations we re left with to solve for a and T: -f k - W 1 sin 30.0 o + T = m 1 a T - W 2 = -m 2 a è T = W 2 - m 2 a Substituting for T in the first equation: -f k - W 1 sin 30.0 o + W 2 - m 2 a = m 1 a a = (-f k - W 1 sin 30.0 o + W 2 )/(m 1 + m 2 ) = ( (8.00)(9.80)(0.500) + (22.0)(9.80))/( ) = 5.20 m/s 2 T = W 2 - m 2 a = (22.0)(9.80) - (22.0)(5.20) = 101 N

Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4. Forces and Newton s Laws of Motion. continued Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting