1. Chapter 6 Practice Problems

2. Equations Sin θ = opposite side hypotenuse Cos θ = adjacent side hypotenuse Tan θ = opposite side adjacent side

3. Equations v R 2 = v p 2 + v w 2 R 2 = A 2 + B 2 F v = F sin θ F h = F cos θ F net 2 = F netx 2 + F nety 2 A + B + W = 0

4. F ll = W Sin θ F l = W cos θ F f = μF N

5. Problem 1 Finding a resultant velocity. An airplane flying toward 0o at 90 km /h is being blown toward 90o at 50 km/h. What is the resultant velocity of the plane. Given: v p = 90 km /h at 0 o v w = 50 km/h at 90o tan Θ = side opposite /side adjacent Unknown ; resultant velocity v r

6. Problem 1 Basic equation: R 2 = A 2 + B 2 or v R 2 = v p 2 + v w 2

7. Problem 1 Answer: v R = 103 km/ h at 29 o

8. Problem 1 Solution: V R 2 = (90 km/h) 2 + (50 km/h) 2 = sq root 1.06 X 10 4 (km/h )2 v R 2 = 103 km /h Tan θ = 50 km/h 90 km/h Tan θ -1 =.556 = 29 o

9. Problem 2 – Resolving a Velocity Vector into its Components – A wind with a velocity of 40 km/h blows toward 30 o. – A. What is the component of the wind’s velocity toward 90 o – B. What is the component of the wind’s velocity toward 0 o

10. Problem 2 Given: v= 40.0 km/h, 30.0 o Unknown: v 90, v 0 Solution: Toward 0 o and 90 o are positive. Angles are measured from 0 o. To find the component toward 90 o use the following: sin 30 o = v 90 /v then v 90 = 20 km/h at 90 o

11. Problem 2 From 0 o cos 30 o =v o /v V 0 = 34.6 km/h at 0.0 o

12. Problem 3 Adding Non perpendicular Vectors Two ropes(F 1 = 12.0 N at 10.0 o ) (F2= 8.0 N at 120 o ) are pulling on a log. What is the net force on the log? Given: F 1 = 12.0 N at 10.0 o F 2 = 8.0 N at 120 o Unknown: F net

13. Problem 3 Find the perpendicular components of each force.(figure 6-12) F 1x =(12 N) cos 30 o = 11.8 N F 1y = (12 N) sin 30 o = 2. N F 2x = ( 8.0 N) cos 120 o = -4.0 N F 2y = (8.0 N) sin 120 o = 6.9 N

14. Problem 3 Sum x and y components F net x = F 1x + F 2x = 11.8 N + -4.0 N = 7.8 N F nety = F 1y + F 2y = 2.0 N + 6.9 N = 8.9 N Find the magnitude of the net force F net = sq root F x 2 +F y 2 = sq root of (7.8 N) 2 + (8.9 N ) 2 = 11 N

15. Problem 3 Find the angle of the force Tan θ = 1.14 Θ = 49 o

16. Problem 4 Finding Forces when Components are Known A sign that weighs 168 N is supported by ropes a and b (figure6-16)that make 22.5 o angles with the horizontal. The sign is not moving. What forces do the ropes exert on the sign?

17. Problem 4 Given: The sign is in equilibrium Weight = W = 168 N (down) Angles ropes make with horizontal is 22.5 o Unknowns: Force of rope a is A Force of rope b is B

18. Problem 4 Basic equations: In equilibrium net force is 0 A + B + W = 0 F h = F Cos θ F v = F Sin θ

19. Problem 4 Since W is down the direction of A + B is up Since the sum of A + B has no horizontal components. Therefore A h and B h have equal magnitude Now, A h = A Cos 22.5 o and B h = B cos 22.5 o A v + B v = 168 N Since A = B then Av + Bv

20. Problem 4 Thus Av = Bv = ½ (168 N ) = 84 N A = A v sin 22.5o = 220 N B = A = 220 N

21. Problem 5 Finding F l and F ll A trunk weighing 562 N is resting on a plane inclined at 30 o from the horizontal (Figure 6-18 ) Find the components of the weight parallel and perpendicular to the plane.

22. Problem 5 Given: W = 562 N Θ = 30 o Unknown: F l F ll

23. Problem 5 Solution: Resolve the weight into components perpendicular and parallel to plane F ll = W sin θ F l = W Cos θ

24. Problem 5 F ll = (+562 N)( sin 30.0 o ) = +(562 N) (0.500) = +281 N F l = + (562 N)(cos 30 o ) = + (562 N) (.866 ) = + 487 N

25. Problem 6 Finding acceleration down a plane The 562 N trunk is on a frictionless plane inclined at 30o from the horizontal. Find the acceleration of the trunk. What is its direction?

26. Problem 7 Finding the Coefficient of Static Friction