Help Desk: 9:00-5:00 Monday-Thursday, 9:00-noon Friday, in the lobby of MPHY. SI (Supplemental Instructor): Thomas Leyden (thomasleyden@tamu.edu) 7:00-8:00pm, Sunday/Tuesday/Thursday, MPHY 333
Chapter 5 Applications of Newton s Laws To draw free-body diagrams and analyzing forces on individual objects. To solve for unknown quantities using Newton's 2 nd law. To understand kinetic and static friction forces acting on an object. To use Hooke's law to relate the restoring force of a spring to the amount of stretching or compression of a spring.
5.1 Equilibrium of a Particle If σ റF = 0 or σ F x = 0 σ F y = 0 y C then റa = 0 or a x = 0 v x = constant a y = 0 v y = constant B T B T C θ T A x See notes for Chapter 4. m A
Newton s Second Law: in vector form 5.2 Applications of Newton s Second Law σ റF = m റa in component form σ F x = ma x σ F y = ma y Strategy for Solving Newton s Law Problems A Review of Problem-Solving Strategy Isolate the bodies in a system. Analyze all the forces acting on each body and draw one free-body diagram for each body. Apply Newton s Third Law to identify action-reaction force pairs. Based on the free-body diagram, set up a most convenient x-y coordinate system. Break each force into components using this coordinate system. For each body, sum up all the x-components of the forces to an equation: σ F x = ma x. For each body, sum up all the y-components of the forces to an equation: σ F y = ma y. Use these equations to solve for unknown quantities.
5.3 Contact Force and Friction Friction force exits at the contact surfaces of two objects. There are two regions of friction: when an object is sliding with respect to another kinetic-friction force when there is no relative motion static-friction force Kinetic Friction Force f k Magnitude: It is directly related to the normal force. f k = μ k n μ k : coefficient of kinetic friction Direction: Parallel to the contact surface Opposite to the direction of relative motion A box being pushed on a table surface force n force റf റF W Static Friction Force f s Magnitude: It depends. It is not fixed. Maximum value f s,max = μ s n μ s : coefficient of static friction Direction: Parallel to the contact surface Opposed to the tendency to change the state of motion
f s = T before breaking loose Distinguishing Static and Kinetic Friction Forces
Example: Two boxes of masses m 1 and m 2 are pushed together by a horizontal force F to accelerate to the right on a leveled table surface as shown. The coefficient of kinetic friction between either of the boxes and the table surface is μ k. Calculate the acceleration and the normal force between the boxes. F m 1 m 2 m 1 n n 1 n 2 m 2 n y f k1 m 1 g x-axis: F n f k1 = m 1 a f k1 = μ k m 1 g y-axis: n 1 m 1 g = 0 n 1 = m 1 g F f k2 m 2 g x-axis: n f k2 = m 2 a f k2 = μ k m 2 g y-axis: n 2 m 2 g = 0 n 2 = m 2 g F f k1 f k2 = m 2 a + m 1 a = (m 1 + m 2 )a a = (F f k1 f k2 )/(m 1 +m 2 ) = (F μ k m 1 g μ k m 2 g)/(m 1 +m 2 ) n = f k2 +m 2 a = μ k m 2 g +m 2 a = [m 2 /(m 1 + m 2 )]F x
Your Attempt to Lift a Box Consider a box of weight W resting on a leveled floor. You bend down and apply an upward force F trying to lift the box up but you are unsuccessful. Which of the following is the correct relationship between W, F, and the normal force N that the floor exerts on the box? (A) F + N < W (B) F + N > W (C) F + N = W (D) F > N (E) F < N റF
Tension Force in a Chain Problem 4.32 A uniform 10.0 kg chain 2.00 m long supports a 50.0 kg chandelier from the ceiling of a large public building. The tension force at the middle chain should be: (A) (B) (C) (D) (E) 98.0 N 490 N 539 N 49.0 N none of the above
Tension Force in a Chain Problem 4.38 Your mass is 100.0 kg. You are initially standing at rest on a scale inside an elevator. This scale measures force in units of Newton. Then, the elevator starts to descend at a constant acceleration of magnitude 1.000 m/s 2. What should be the reading on the scale while the elevator is descending at this acceleration? (A) (B) (C) (D) 980 N 1080 N 880 N none of the above
Magnitude and Direction of Static Friction Force Example: Box A of mass m A is stacked on top of Box B of mass m B. The friction coefficient between the contacting surfaces is sufficiently large, as such, there is no slipping between the two boxes. Box B is on a horizontal frictionless table surface. Find the magnitude and the direction of the friction force Box B exerts on Box A when they are moving without slipping to the right with a constant velocity V 0. The friction force that Box B exerts on Box A is: (A) f k = μ k n = μ k m A g, to the right (B) f k = μ k n = μ k m A g, to the left (C) f s = μ s n = μ s m A g, to the right (D) f s = μ s n = μ s m A g, to the left (E) 0 A B V 0
Magnitude and Direction of Static Friction Force Example: Box A of mass m A is stacked on top of Box B of mass m B. The friction coefficient between the contacting surfaces is sufficiently large, as such, there is no slipping between the two boxes. Box B is on a horizontal frictionless table surface. Due to a force applied on Box B as shown, the two boxes are moving without slipping with a constant acceleration a 0 pointing to the right. Find the magnitude and the direction of the friction force Box B exerts on Box A. The friction force that Box B exerts on Box A is: (A) f = μ k n = μ k m A g, to the right (B) f = μ s n = μ s m A g, to the left (C) f = m A a 0, to the right (D) f = m A a 0, to the left (E) 0 B A റF
Magnitude and Direction of Static Friction Force Example: Box A of mass m A is stacked on top of Box B of mass m B. The friction coefficient between the contacting surfaces is sufficiently large, as such, there is no slipping between the two boxes. Box B is on a horizontal frictionless table surface. Due to a force applied on Box B as shown, the two boxes are moving without slipping with a constant acceleration a 0 pointing to the right. Find the magnitude and the direction of the friction force Box A exerts on Box B. The friction force that Box A exerts on Box B is: (A) f = m B a 0, to the right (B) f = m B a 0, to the left (C) f = m A a 0, to the right (D) f = m A a 0, to the left (E) 0 B A റF
Additional Examples A box having an initial velocity 10 m/s enters a region of a leveled rough surface and slides a distance of 20 m before it reaches a stop. Assume that the acceleration due to friction is a constant. Calculate the coefficient of kinetic friction. Solution: Initial velocity v 1 = 10 m/s; Final velocity v 2 = 0.0 m/s What is the acceleration? v 2 2 v 1 2 = 2ad a = (v 2 2 v 1 2 )/2d = v 1 2 /2d = 2.5 m/s 2 Free-body diagram and Newton s Second Law y y-axis: n mg = 0 x-axis: f k = ma and f k = m k n m k = f k /n = ma/mg = a/g = 2.5/9.8 = 0.26 f k n mg റv x
Example: Box of mass m is pushed up an incline by a force F parallel to the incline. The coefficient of kinetic friction between the box and the incline is μ k. y f k n F x Given: F, m, μ k, and q Find: n and a Solutions: Weight and its components W = mg W x = mgsinq W y = mgcosq x-axis: F - mgsinq - f k = ma f k = μ k n = μ k mgcosq q q mg a = (F - mgsinq - μ k mg)/m y-axis: n mgcosq = 0 n = mgcosq What if some other quantities are given and you are asked to calculate some different quantities? Consider Problem 9 on Term Exam 1 from 2015.
5.4 Elastic Force Spring, Spring Restoring Force, and Hooke s Law Hooke s Law on spring restoring force: In magnitude F spr = k L Direction: Opposed to length change A general expression taking care of the direction F spr = kx restoring force x is measured with respect to the equilibrium position Example 5.14 A vertical spring balance scale stretches 1.00 cm when a 12.0 N weight is hung on it. If the 12.0 N weight is replaced by a 1.50 kg fish, by what amount is the spring stretched? Answer: Spring constant k = F/ L = (12.0 N)/(0.0100 m) = 1200 N/m Stretching = F/k = mg/k = 0.0123 m = 1.23 cm (c) equilibrium at x = 0 Direction of force with stretching Direction of force with compression
Example with massless and non-stretchable rope & massless and frictionless pulley. n T Given: m and M. Answer the questions for the following two cases. mg T Mg (a) If the table top is frictionless. Question 1: what is the condition under which M accelerates downward? Question 2: If M accelerates downward, calculate the acceleration a and the tension in the rope T. (b) Assume the coefficient of friction between m and the table top to be to be μ s = μ k = μ. Question 1: what is the condition under which M accelerates downward? Question 2: If M accelerates downward, calculate the acceleration a and the tension in the rope T.
Example with massless and non-stretchable rope & massless and frictionless pulley. n T T Given: m 1, m 2, and θ. m 1 Answer the questions for the following two cases. θ m 1 g m 2 m 2 g (a) If the incline is frictionless. Question 1: what is the condition under which m 2 accelerates downward? Question 2: If m 2 accelerates downward, calculate the acceleration a and the tension in the rope T. (b) Assume the coefficient of friction between m 1 and the incline to be μ s = μ k = μ. Question 1: what is the condition under which m 2 accelerates downward? Question 2: If m 2 accelerates downward, calculate the acceleration a and the tension in the rope T.