Chapter 6 Practice Problems

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 • vR2 = vp2 + vw2 • R2 = A2 + B2 • Fv = F sin θ • Fh = F cosθ • Fnet2 = Fnetx2 + Fnety2 • A + B + W = 0

4. Fll= W Sin θ • Fl = W cosθ • Ff = μFN

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: vp = 90 km /h at 0o vw= 50 km/h at 90o tanΘ = side opposite /side adjacent • Unknown ; resultant velocity vr

6. Problem 1 • Basic equation: R2 = A2 + B2 or • vR2 = vp2 + vw2

7. Problem 1 • Answer: vR = 103 km/ h at 29o

8. Problem 1 • Solution: • VR2 = (90 km/h)2 + (50 km/h)2 = • sq root 1.06 X 104 (km/h)2 • vR2 = 103 km /h • Tan θ = 50 km/h • 90 km/h • Tan θ -1= .556 = 29o

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

10. Problem 2 • Given: v= 40.0 km/h, 30.0o • Unknown: v90, v0 • Solution: Toward 0o and 90o are positive. Angles are measured from 0o. To find the component toward 90o use the following: • sin 30o = v90/v then v90 = 20 km/h at 90o

11. Problem 2 • From 0ocos 30o =vo/v • V0 = 34.6 km/h at 0.0o

12. Problem 3 • Adding Non perpendicular Vectors • Two ropes(F1 = 12.0 N at 10.0o) (F2= 8.0 N at 120o ) are pulling on a log. What is the net force on the log? • Given: F1 = 12.0 N at 10.0o • F2 = 8.0 N at 120o • Unknown: Fnet

13. Problem 3 • Find the perpendicular components of each force.(figure 6-12) • F1x =(12 N) cos 30o = 11.8 N • F1y = (12 N) sin 30o = 2. N • F2x = ( 8.0 N) cos 120o = -4.0 N • F2y = (8.0 N) sin 120o = 6.9 N

14. Problem 3 • Sum x and y components • Fnet x = F1x + F2x = 11.8 N + -4.0 N = 7.8 N • F nety = F1y + F2y= 2.0 N + 6 .9 N = 8.9 N • Find the magnitude of the net force • F net = sq root Fx2 +Fy2 = 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 = 49o

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 22.5o 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 22.5o • 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 • Fh = F Cos θ • Fv = 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 Ah and Bh have equal magnitude • Now, Ah = A Cos 22.5o and Bh = B cos 22.5o • Av + Bv = 168 N • Since A = B then Av + Bv

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

21. Problem 5 • Finding F l and Fll • A trunk weighing 562 N is resting on a plane inclined at 30o 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 = 30o • Unknown: • Fl • F ll

23. Problem 5 • Solution: • Resolve the weight into components perpendicular and parallel to plane • Fll= W sin θ Fl= W Cos θ

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