Aim: How can we describe Newton’s 1 st Law of Motion?

1. Aim: How can we describe Newton’s 1st Law of Motion? HW #4 Do Now: Resolve the following vector into its horizontal and vertical components: Answer Key HW 3 200 N x-component x = (200 N)cos37° x = 200(4/5) x = 160 N y-component y = (200 N)sin37° y = 200(3/5) y = 120 N 37° No Calculator!!

2. Newton’s 1st Law of Motion Law of Inertia If all the forces acting on an object are balanced: a. an object at rest remains at rest b. an object in motion will remain in motion with a constant velocity ΣFx = 0 ΣFy = 0

3. Inertia – the ability of an object to resist a change in motion • Objects at rest want to remain that way • Objects in motion want to remain that way • Inertia is dependent only on mass • (the more mass, the more inertia)

4. Three objects can only move along a straight, level path. The graphs below show the position d of each of the objects plotted as a function of time t. No Calculator **1 minute** • 1. The sum of the forces on the object is zero in which of the cases? • II only • III only • I and II only • I and III only • I, II, and III Graph I – constant velocity Graph II – At rest Graph III - Accelerating

5. 2. Three forces act on an object. If the object is in translational equilibrium, which of the following must be true? • I. The vector sum of the three forces must equal zero. • II. The magnitudes of the three forces must be equal. • III. All three forces must be parallel. • I only • II only • I and III only • II and III only • I, II, and III II – 3 forces in the same direction all add up to a resultant force, yielding an unbalanced force III – same reasoning as II No Calculator **1 minute**

6. The key to the problems is the free-body diagram! FN (also referred to as the apparent weight of the object in certain problems) FF F For objects at rest or moving with a constant velocity: F = Ff FN = Fg Fg Fg = mg

7. Coefficient of Friction Represented as µ FF = µFN µs = coefficient of static friction (objects at rest) µk = coefficient of kinetic friction (objects in motion) µ is a unitless number between 0 (no friction) and 1 (100% friction) Snow and ice have a low µ Rubber has a high µ

8. FN How can we solve for F║ and F┴ Ff FII Mathematically, these θ’s are equal θ F θ Fg = mg

9. For objects at rest or moving with a constant velocity on an incline: Fll = Ff FN = F┴

10. No Calculator **2 minutes ** A plane 5 meters in length is inclined at an angle of 37°, as shown. A block of weight 20 newtons is placed at the top of the plane and allowed to slide down. 3. The mass of the block is most nearly(A) 1.0 kg (B) 1.2 kg (C) 1.6 kg (D) 2.0 kg (E) 2.5 kg 4. The magnitude of the normal force exerted on the block by the plane is most nearly(A) 10 N (B) 12 N (C) 16 N (D) 20 N (E) 33 N Fg = mg 20 N = m(10 m/s2) m = 2 kg FN = F┴ FN = Fgcosθ FN = (20 N)(4/5) FN = 16 N

11. Resolve this into its components F Fcosθ θ Fsinθ

12. No Calculator **1 minute** • 5. A ball of mass m is suspended from two strings of unequal length as shown above. The tensions T1 and T2 in the strings must satisfy which of the following relations? • Tl = T2 • T1 > T2 • T1 < T2 • Tl + T2 = mg • T1‑T2 = mg T2 has a larger vertical component and therefore is supporting more of the weight of mass m

13. No Calculator **1 min 15 sec** 6. A push broom of mass m is pushed across a rough horizontal floor by a force of magnitude T directed at angle  as shown above. The coefficient of friction between the broom and the floor is . The frictional force on the broom has magnitude (A) (mg + Tsin) (B) (mg-Tsin) (C) (mg+ Tcos) (D) (mg-Tcos) (E) mg FF = μFN FF = μ(mg + Tsinθ) ΣFy = 0 FN = mg + Tsinθ FN FF Tcosθ mg Tsinθ

14. No Calculator **1 min 15 sec** 7. A block of weight W is pulled along a horizontal surface at constant speed v by a force F, which acts at an angle of  with the horizontal, as shown above. The normal force exerted on the block by the surface has magnitude (A) W ‑ F cos (B) W‑Fsin (C) W (D) W + Fsin (E) W + Fcos FN Fsinθ ΣFy = 0 FN + Fsinθ – W = 0 FN + Fsinθ = W FN = W - Fsinθ FF Fcosθ W

15. No Calculator **1 min 15 sec** Question #8 FF = μFN FF = μ(15 N + mg) FF = 0.2(15 + 20) FF = 0.2(35) FF = 7 N FN ΣFy = 0 FN – 15 N – mg = 0 FN = 15 N + mg 15 N mg

16. No Calculator **1 min 15 sec** • 9. When an object of weight W is suspended from the center of a massless string as shown above. The tension at any point in the string is • 2Wcos • ½Wcos • (C) Wcos • (D) W/(2cos) • (E) W/cos ΣFx = 0 Tsinθ – Tsinθ = 0 X – components cancel out ΣFy = 0 Tcosθ + Tcosθ – W = 0 2 Tcosθ = W T = W/(2 cosθ) Tcosθ Tcosθ Tsinθ Tsinθ W

17. No Calculator **1 min 15 sec** • 10. A uniform rope of weight 50 newtons hangs from a hook as shown above. A box of weight 100 newtons hangs from the rope. What is the tension in the rope? • 50 N throughout the rope • 75 N throughout the rope • 100 N throughout the rope • 150 N throughout the rope • It varies from 100 N at the bottom of the rope to 150 N at the top. Top Bottom ΣF = 0 T = 100 N ΣF = 0 T = 100N + 50 N T = 150 N T T 100 N 100 N 50 N

18. Calculator allowed **2.5 minutes** FN 11. A box of uniform density weighing 100 newtons moves in a straight line with constant speed along a horizontal surface. The coefficient of sliding friction is 0.4 and a rope exerts a force F in the direction of motion as shown above. a. On the diagram to the right, draw and identify all the forces on the box. F FF Fg ΣFx = 0 F = FF F = μFN F = μFg F = (0.4)(100 N) F = 40 N b. Calculate the force F exerted by the rope that keeps the box moving with constant speed.

19. Calculator allowed **5 minutes** 12. A 5.0‑kilogram monkey hangs initially at rest from two vines, A and B. as shown above. Each of the vines has length 10 meters and negligible mass. a. On the figure to the right, draw and label all of the forces acting on the monkey. (Do not resolve the forces into components, but do indicate their directions.) TB TA mg b. Determine the tension in vine B while the monkey is at rest.