Application of the Laws of Motion Lecture Notes PDF

Application of the Laws of Motion LECTURER Dr. Edwin Nyirenda and Mr Keith Nsofwa Forces of Friction  When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion.  Friction is a force that opposes relative motion be...

Application of the Laws of Motion LECTURER Dr. Edwin Nyirenda and Mr Keith Nsofwa Forces of Friction  When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion.  Friction is a force that opposes relative motion between systems in contact.  We will be concerned with two types of frictional force  Force of static friction: fs  Force of kinetic friction: fk  Direction: opposite the direction of the intended motion  If moving: in direction opposite the velocity  If stationary, in direction of the vector sum of other forces Friction cont.  Magnitude: Friction is proportional to the normal force  Static friction: Ff = F  μsN  Kinetic friction: Ff = μkN  μ is the coefficient of friction  The coefficients of friction are nearly independent of the area of contact (why?)  Static Friction  If two systems are in contact and stationary relative to one another, then the friction between them is called static friction.  If 𝑭 increases, so does 𝒇𝒔  If 𝑭 decreases, so does 𝒇𝒔  ƒs  µs N  Remember, the equality holds when the surfaces are on the verge of slipping Kinetic Friction  If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction.  The force of kinetic friction acts when the object is in motion  Although µk can vary with speed, we shall neglect any such variations  ƒk = µk N  Friction cont.  For small applied forces, the magnitude of the force of static friction equals the magnitude of the applied force.  When the magnitude of the applied force exceeds the magnitude of the maximum force of static friction, the trash can breaks free and accelerates to the right. Example 1  A 20.0-kg crate is at rest on a floor as shown. The coefficient of static friction between the crate and floor is 0.700 and the coefficient of kinetic friction is 0.600. A horizontal force P is applied to the crate. Find the force of friction if (a) P = 20.0 N, (b) P = 30.0 N, (c) P = 120.0 N, and (d) P = 180.0 N  Example 1 cont.  Solution  Newton’s fist law gives F y 0 N W  0 N W f s   s N  0.700  20.0kg  9.81m / s 2  137 N  As long as P is less than 137 N, the force of static friction keeps the crate stationary and 𝑓 = 𝑷. Thus,  (a) 𝑓 = 20.0 𝑁, (b) 𝑓 = 30.0 𝑁, and (c) 𝑓 = 120.0 𝑁.  (d) If P = 180.0 N, the applied force is greater than the maximum force of static friction (137 N), so the crate can no longer remain at rest. Once the crate is in motion, kinetic friction acts. Then  Example 1 cont.  The friction and acceleration will be f k   k N  0.600  196 N  118 N F x  ma P  f k  ma P  f k 180.0 N  118 N a   3.10m / s 2 m 20.0kg  Example 2  suppose you move the crate by pulling upward on the rope at an angle of 30° above the horizontal. a) How hard must you pull to keep it moving with constant velocity? Assume that 𝜇 = 0.40. b) Determine the normal force n c) The friction force  Example 2 cont.  The crate is in equilibrium because its velocity is constant(𝑎 = 0𝑚/𝑠 ) F x 0 T cos 30  (  f k )  0 T cos 30  f k F y 0 T sin 30  n  (  w)  0 n  w  T sin 30 Example 2 cont. BUT f k  k n  k ( w  T sin 30) but T cos 30   k ( w  T sin 30) k w T  188 N cos 30  k sin 30  Determine the normal force n n  w  T sin 30  500 N  (188 N ) sin 30  406 N f k  k n  0.4  406 N  162.4 N  EXAMPLE 3  Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down?  EXAMPLE 3 CONT.  Newton 2nd law: F x  mg sin   s N  0 F y  N  mg cos  0  Then N  mg cos F y  mg sin  s mg cos  0 tan    s  0.350  So   tan 1 (0.350)  19.3  Example 4  The snowboarder glides down a slope that is inclined at θ = 13° to the horizontal. The coefficient of kinetic friction between the board and the snow is 𝜇 = 0.20. What is the acceleration of the snowboarder?  Example 4 cont.  Solution  We can now apply Newton’s second law to the snowboarder. F x  max mg sin   k N  max F y 0 N  mg cos   0 N  mg cos   From the second equation, N = mg cos θ. Upon substituting this into the first equation, we find ax  g (sin    k cos  )  g (sin13  0.520 cos13)  0.29m / s 2  Task (solved)  A block of mass 𝑚 on a rough, horizontal surface is connected to a ball of mass 𝑚 by a lightweight cord over a lightweight, frictionless pulley as shown in figure. A force of magnitude F at an angle 𝜃 with the horizontal is applied to the block as shown and the block slides to the right. The coefficient of kinetic friction between the block and surface is μk. Find the magnitude of acceleration of the two objects. Example 2 cont.  m1: F x  F cos  f k  T  m1a x  m1a F y  N  F sin   m1 g  0  m2: F y  T  m2 g  m2 a y  m2 a T  m2 (a  g ) N  m1 g  F sin f k   k N   k ( m1 g  F sin ) F cos   k ( m1 g  F sin )  m2 (a  g )  m1a F (cos   k sin  )  (m2   k m1 ) g a m1  m2 Review and questions