1. Force (Weight) (Tension) Friction Force MIDTERM on 10/06/10 7:15 to 9:15 pm  Bentley 236  2008 midterm posted for practice.  Help sessions Mo, Tu 6-9 pm.  2-side hand written sheet for equation/notes.  18 MC, 3 long problems CAPA 4a due this Friday (10/01/10) at 11:59 PM CAPA 4b due next Tuesday (10/05/10) at 11:59 PM No quiz on this Friday. Quiz 4 on 10/05/10, next week Tuesday 10/01/2010

2. A net force of 250N is exerted to the right on a large box of mass 50kg. What is the acceleration of the box? (1) 0.2 m/s 2 to the right(2) 0.2 m/s 2 to the left (3) 1.25 m/s 2 to the right(4) 1.25 m/s 2 to the left (5) 5.0 m/s 2 to the right(6) 5.0 m/s 2 to the left (7) 12.5 m/s 2 to the right(8) 12.5 m/s 2 to the left Acceleration same direction as the force. Clicker

3. You and a friend are sliding a large 100-kg box across the floor. Your friend pulls to the right with a force of 250N. You push to the right with a force of 300N. The frictional force of the floor opposes the motion with a force of 500N. What is the acceleration of the box? Example 1 250N 300N 500N 100 kg

5. Connections - Gravity Abell 1060 / Hydra

6. Newton’s Law of Gravitation m 1 m 2 r Gravitational constant (G), mass of the Earth (M E ), and radius of the Earth (r E ) are constants. So we can determine the gravitational acceleration due to gravity (g) as 9.81 m/s 2.

7. A force of 300 N pointing towards East is applied to a block of 100 kg. Another force of 300 N is applied towards North. If the friction is negligible, find the acceleration of the block. Example 2 300N Net Force 100 kg

8. A group is pushing a box of 100 kg. A pushes to the East with 100 N, B to the North-East with 200 N, C to the South with 150 N, and D to the North-West with 100 N. What will be the net acceleration of the box? Example 3 100N 150N 200N 100 kg A B C 100N D

9. A FedEx employee pushes a 50-kg box initially at rest on the floor to the right. 10 s later the box is moving with a speed of 1.5 m/s. Find the net force used by the employee. 50 kg Example 4 (1) V = V 0 + at (2)  x = ½ (V 0 + V) t (3)  x = V 0 t + ½ at 2 (4) V 2 = V 0 2 + 2 a  x

10. A FedEx employee pushes a 50-kg box initially at rest on the floor to the right. 10 s later the box is displaced 20 m from the initial position. Find the net force used by the employee. 50 kg Example 5 (1) V = V 0 + at (2)  x = ½ (V 0 + V) t (3)  x = V 0 t + ½ at 2 (4) V 2 = V 0 2 + 2 a  x

11.  W = m g CLICKER! The Moon’s gravitational acceleration is six times less than that of the Earth. What is the weight of an astronaut on Moon whose mass is 80 kg? Gravitational acceleration on Earth is 9.81 m/s 2. a)262 N b)131 N c)49 N d)0.02 N = 80 x (9.81)/6 = 130.8 N Example 6

12. m Moon m Earth r The mass of the Moon is 7.36 E22 kg and the mass of the Earth is 5.9742E24 kg. If the distance between the Earth and Moon is 3.84E8 m, find the gravitational force between them. Example 7

13. Lab – Data inexact Various slopes – max and min from steepest and shallowest reasonable lines Draw Straight lines, not “Connect the Dots” Acceleration down an incline: If θ = 0 (Horizontal), no acceleration (a=0) If θ = 90 (Vertical), a=g If in between, a = g sinθ x = v 0 t + ½at 2 which reduces to x = ½at 2 if v 0 = 0 m/s Lab

14. The graph shows the speed of a car as a function of time. Selected the correct statement. 1). F B = F C 2). F A > F E = F D 3). F E > F B > F C Larger the slope  higher acceleration (a =  v/t)  larger force. Clicker

15. Normal Force The force exerted by the surface. It is always perpendicular to the surface. (The object must be located on a surface.) Normal force has an equal magnitude and opposite indirection with the weight or weight component.

16. Free Body Diagram The drawing of forces acting on a body. - weight (W) downward. - on a surface; normal force (F N ) perpendicular to the surface. - ropes; add tensions (T). W FNFN W T1T1 T2T2

17. Equilibrium The net force acting on a body is zero. (The forces cancel out. Therefore, an object at rest remains at rest, and an object in motion continues to travel with a constant velocity).

18. A large crate weighs 100N. You push straight down on the top of the crate with a force of 75N. What is the force with which the floor pushes up on the crate? (1)25N (2)75N (3) 100 N (4) 125N (5) 175N ΣF Y = 0 F FLOOR – 100 N – 75 N = 0 F FLOOR = 100N + 75N FGFG F YOU F FLOOR What if a Y wasn't zero? Just plug in value. Clicker

19. What is the magnitude of the upward force (normal force) of the table on the box? (1) 100N (2) 100N + F cosθ (3) 100N - F cosθ (4) 100N + F sinθ (5) 100N - F sinθ (6) F cosθ (7) F sinθ ΣF Y = 0 F N – 100 N – Fsinθ = 0 F N = 100 N + Fsinθ FYFY FxFx +y +x Clicker

20. You are on an elevator which is accelerating upward. How does the normal force (F N ) compare to your weight (W)? 1.normal force > weight 2.normal force = weight 3.normal force < weight Clicker

21. Apparent Weight We can increase or decrease the weight by adding or subtracting additional vertical force on the object. (lift is stationary) (lift moves upward) (lift moves downward ) W = F N W app > F N W app < F N

22. (lift is stationary) (lift moves upward) (lift moves downward ) W = F N W app > F N W app < F N You are on an elevator which is accelerating upward at a rate of 2.00m/s 2. Your mass is 80.0kg, and the elevator has a mass of 500.kg. A cable is used to pull the elevator upward. What is the force of the cable on the elevator? Example 8

23. Frictional Forces

24. You push a crate on the floor. The crate starts moving at 100N force. If the coefficient of fraction is 0.15, find the mass of the crate. Example 9

25. Incline x y   W FNFN f FxFx

26. Multiple Masses Tied by Ropes T1T1 T2T2 > T 2 = (m 1 + m 2 ) a T 1 = m 1 a

27. You are standing still on ice (consider this to be frictionless). Your friend (mass=60kg) pushes you (mass=80kg). Your acceleration is +1.0m/s 2. What is your friend's acceleration? (1) 0 m/s 2 (2) +0.75 m/s 2 (3) +1.0 m/s 2 (4) +1.33 m/s 2 (5) -0.75 m/s 2 (6) -1.0 m/s 2 (7) -1.33 m/s 2 Σ Horizontal Forces on You = ma YOU F FRIEND ON YOU = (80kg)*(+1m/s) = +80 N You exert same magnitude force in opposite direction on them. ΣHorizontal Forces on Friend = ma FRIEND -80N = (60kg)a FRIEND a FRIEND = - 1.33 m/s 2 (opposite direction) Clicker