**Midterm Exam #1** • First midterm is this Friday, January 31 (75 minutes!!!) • Units 1-5: 1D kinematics – Forces and FBDs • Things to bring: Pencil, calculator, 1 sheet of notes (8.5 X 11 in., front & back), UofU ID card!!! • No makeups!!! (Unless you are unavailable due to University/military business or documented medical emergency) • Cheating: You will receive a zero and referred to the University administration – may result in an “E” for the course and expulsion • Multiple choice: 130 points total • Locations:

**Multiple Choice with Partial Credit** h=3630 km Earth RE=6370 km Space Telescope 6 points

**Multiple Choice with Partial Credit** h=3630 km Earth RE=6370 km Space Telescope 3 points

**Multiple Choice with Partial Credit** h=3630 km Earth RE=6370 km Space Telescope 2 points

**Classical Mechanics Lecture 6: Friction** Today's Concept: Friction

**Problem of the Day** A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car’s tires and the road is μs. The coefficient of kinetic friction is μk. What is the maximum speed of the car before it starts to skid?

**y** x Stuff you asked about: 90 90-θ θ 90 mg cos(θ) θ mg mg sin(θ) I struggled with the trigonometry on the box sliding down the ramp. i thought that this was a good lecture, very informative, but could we have an example, just one little example, that doesn't involve a box? this lecture was very clear though.

**Friction**

**CheckPoint: Box in Accelerating Truck** μS a A B C A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding around as the truck drives. If the truck moves with constant acceleration to the left as shown, which of the following diagrams best describes the static frictional force acting on the box:

**Revote: Box in accelerating truck** μS a A B C A) The friction is causing the box to accelerate in the same direction as the truck. B) The force of friction always acts in the opposite direction of acceleration. If the truck moves with constant acceleration to the left as shown, which of the following diagrams best describes the static frictional force acting on the box:

**ACT!** A) Mg B) μMg C) T D) 0 T Since acceleration is zero, the net force MUST be zero also! A flower pot of mass Msits on a horizontal table. A horizontal string having tension Tapplies a force on the pot, but static friction between the pot and the table keeps the pot from moving. What is the magnitude of the net force acting on the pot?

**CheckPoint: Box on Table** A box of mass Msits on a horizontal table. A horizontal string having tensionTapplies a force on the box, but static friction between the box and the table keeps the box from moving. What is the magnitude of the static frictional force acting on the box? A) Mg B) μsMg C) T D) 0 T M Apply Newton’s 2nd Law:

**T** M CheckPoint: Box on Table A box of mass Msits on a horizontal table. A horizontal string having tensionT applies a force on the box, but static friction between the box and the table keeps the box from moving. What is the magnitude of the static frictional force acting on the box? Ffdepends on weight μsdoes not!!! • What happens if I pull harder? • As long as v = 0: The static friction force will continue to match T… • …until Ff = μs N. After that, Ff can’t increase any more: • T > Ff and the box will accelerate!

**Problem of the Day** A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car’s tires and the road is μs. The coefficient of kinetic friction is μk. What is the maximum speed of the car before it starts to skid?

**ACT!** A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car’s tires and the road is μs. The coefficient of kinetic friction is μk. What is the maximum speed of the car before it starts to skid? A) B) C) D)

**ACT!**

**Problem of the Day** A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is μk. What is the tension in the string? m2 g m1

**CheckPoint: Blocks and Pulley** m2 m2 g Case 2 (With Friction) g Case 1 (No Friction) m1 m1 In which case is the tension in the string larger? A) Case 1 B) Case 2 C) Same 69% of you got this right A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table.

**ACT: Blocks and Pulley, I ** a is positive when m1 accelerates downward! m2 Case 1 (No Friction) g m1 What is the tension in the string for Case 1? A) T = 0 B) T = m1gC) 0 <T < m1g A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table.

**ACT: Blocks and Pulley, II** m2 m2 g Case 1 (No Friction) Case 2 (With Friction) g m1 m1 In which case is the acceleration of the blocks larger? A) Case 1B) Case 2C) Same A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table.

**CheckPoint revisited: Blocks and Pulley** m2 m2 g Case 1 (No Friction) Case 2 (With Friction) g m1 m1 One of your answers: B) M1 will not be accelerating as fast in case two than in case one because there is the extra force of friction acting against mass 1. Since the acceleration is smaller in case two, there has to be more of a force acting against gravity, the only other possible force is Tension. In which case is the tension in the string larger? A)Case 1B)Case 2C)Same

**Problem of the Day** A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is μk. What is the tension in the string? m2 g m1

**m2** 1) FBD N T m2 g f T m1 m1 m2g m1g

**m2** 1) FBD 2) ΣF=ma N T m2 g f T m1 m1 m2g N = m2g m1g m1g – T = m1a T – μ m2g = m2a add m1g – μ m2g = m1a + m2a m1g – μ m2g a = m1 + m2

**m2** 1) FBD 2) ΣF=ma N T m2 g f T m1 m1 m2g m1g m1g – T = m1a m1g – μ m2g T = m1g – m1a a = m1 + m2 T is smaller when a is bigger