Chapter 7 Forces in Two Dimensions

1. Chapter 7Forces in Two Dimensions A 50 kg sign is supported in a motionless position supported by two ropes that make a 40º angle with the horizontal What is the Tension in each rope?

2. FA FB 40° 40° Fg FB 50° 490 N 80° 50° FA Chapter 7Forces in Two Dimensions Draw a free body diagram The force of gravity is F=ma or F= (50kg)(9.8 m/s2)=490 N = 490 N Since the system is in equilibrium, the sum of the vectors is zero and by the head tail method FA = FB = 381 N

3. Fx Fy Fg Fy Fg Fx Chapter 7Forces in Two Dimensions Motion along inclined planes By vector addition and similar triangles

4. Fx Fy 300 N Fy 300 N Fx Chapter 7Forces in Two Dimensions A trunk weighing 300 N is placed on a plane and starts to slide when inclined at 30°.Find the parallel and perpendicular components of the force and the coefficient of friction.

5. Fx Fy 300 N Fy 300 N Fx Chapter 7Forces in Two Dimensions Fx=F sin 30°=300 N *.5 Fx = 150 N Fy=F cos 30°=300 N *.866 Fy = 260 N Check (150 N)2+(260 N)2 = (300 N)2 or µ = tan ø

6. Chapter 7Forces in Two Dimensions Suppose the crate is moving down the plane and the coefficient of kinetic friction between the surfaces is .10. How fast is the crate going 2 seconds after starting from rest?

7. Ff Fx Fy 300 N Chapter 7Forces in Two Dimensions y direction FN= mg cos Ø x direction Fnet = Fx - Ff Fnet= Fx-mkFN Applet Fnet= 150 N- .10(260 N) = 124 N F = ma 124 = (300N/9.8 m/s2) a a = 4.0 m/s2 vf = vi + at vf = 0 + 4.0 m/s2*2 s = 8 m/s

9. Projectiles Launched Horizontally

11. 20 m/s 490 m A projectile is thrown horizontally at 20 m/s from the top of a cliff 490 m high. • How long is the object in the air? • How far from the base of the cliff does it land? • How fast is it moving the instant it hits the ground? x0 = 0 vx0 = 20 m/s y0 = -490 m ay = -9.8 m/s2 b. d = vt d = (20m/s)(10s) d = 200 m • d = vit + ½ at2 • solve for t • t = c. vf = vi+at vf = 0 + (-9.8m/s2)(10s) vf = -98 m/s t = 10 s

12. 5 m/s 19.6 m A projectile is thrown horizontally at 5 m/s from the top of a cliff 19.6 m high. • How long is the object in the air? • How far from the base of the cliff does it land? • How fast is it moving the instant it hits the ground? x0 = 0 vx0 = 5 m/s y0 = -19.6 m ay = -9.8 m/s2 b. d = vt d = (5m/s)(2s) d = 10 m • d = vit + ½ at2 • solve for t • t = c. vf = vi+at vf = 0 + (-9.8m/s2)(2s) vf = -19.6 m/s t = 2 s

13. 60º vy vx Projectiles Launched at an Angle A ball is thrown with an initial velocity of 5 m/s at an angle of 60º with the horizon at ground level. vx = v cos Ø = 5 cos 60º= 2.5 m/s vy = v sin Ø = 5 sin 60º = 4.33 m/s 5m/s • How long is the ball in the air? • How high does it go? • How far does it go? 60º x0=0 y0 = 0 v= 5 m/s  = 60º a = -9.8 m/s2

14. 53º vy vx Projectiles Launched at an Angle A projectile is thrown with an initial velocity of 98 m/s at an angle of 53º with the horizon at ground level. vx = v cos Ø = 98 cos 53º= 58.98 m/s vy = v sin Ø = 98 sin 53º = 78.27 m/s 98 m/s • How long is the ball in the air? • How high does it go? • How far does it go? 53º x0=0 y0 = 0 v= 98 m/s = 53º a = -9.8 m/s2

15. ø vy vx Projectiles Launched at an Angle A projectile is thrown with an initial velocity of v m/s at an angle of øwith the horizon at ground level. vx = v cos Ø vy = v sin Ø v • How long is the ball in the air? • How high does it go? • How far does it go? ø x0=0 y0 = 0 v= v =  a = -9.8 m/s2

16. Projectile Motion

19. What goes faster, the inside of a record or the outside? Animation

20. Fc ac Circular Motion

21. As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal force required for circular motion.

22. As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.

23. As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

24. r

25. A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car. Known Information: m = 900 kg v = 10.0 m/s R = 25.0 m Requested Information: a = ???? Fnet = ???? To determine the acceleration of the car, use the equation a = (v2)/R. The solution is as follows: a = (v2)/R a = ((10.0 m/s)2)/(25.0 m) a = (100 m2/s2)/(25.0 m) a = 4 m/s2 To determine the net force acting upon the car, use the equation F net = m*a. The solution is as follows. Fnet = m*a Fnet = (900 kg)*(4 m/s2) Fnet = 3600 N

26. Joe Brick, the new 95-kg halfback makes a turn on the football field. He sweeps out a path which is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting on Joe. Requested Information: v = ???? a = ???? Fnet = ???? Known Information: m = 95.0 kg R = 12.0 m Traveled 1/4-th of the circumference in 2.1 s To determine the speed of the halfback, use the equation v = d/t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows: v = d/t v = (0.25 * 2 π * R)/t v = (0.25 * 2 * 3.14 * 12.0 m)/(2.1 s) v = 8.97 m/s

27. To determine the acceleration of the halfback, • use the equation a = (v2)/R. The solution is as follows: • a = (v2)/R • a = ((8.97 m/s)2)/(12.0 m) • a = (80.5 m2/s2)/(12.0 m) • a = 6.71 m/s2 • To determine the net force acting upon the halfback, • use the equation Fnet = m*a. The solution is as follows. • Fnet = m*a • Fnet = (95.0 kg)*(6.71 m/s2) • Fnet = 637 N

28. Simple Harmonic Motion Definitions • Period (s) The time required to complete one cycle. • Frequency (hz) The number of cycles per second. • Amplitude (m) The point of maximum displacement from • rest.

29. E D D Distance (meters) C C B B A A Simple Harmonic Motion

30. C D B A A E Velocity (m/s) D B C Simple Harmonic Motion

31. C Acceleration (m/s 2) D B A E A D B C Simple Harmonic Motion

32. Simple Harmonic Motion