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PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS

PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS. EQUILIBRIUM & EQUILIBRIANT FORCES A. GRAPHICAL ADDITION OF VECTORS 1. PYTHAGOREAN THEOREM R 2 = A 2 + B 2 2. LAW OF COSINES R 2 = A 2 + B 2 – 2ABcosO 3 . VECTOR COMPONENTS ARE “HEAD TO TAIL”

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PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS

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  1. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS • EQUILIBRIUM & EQUILIBRIANT FORCES • A. GRAPHICAL ADDITION OF VECTORS • 1. PYTHAGOREAN THEOREM • R2 = A2 + B2 • 2. LAW OF COSINES • R2 = A2 + B2 – 2ABcosO • 3. VECTOR COMPONENTS ARE “HEAD TO TAIL” • 4. VECTOR ADDITION (X AND Y COMPONENTS) • B. EQUILIBRIANT • 1.EQUAL AND OPPOSITE OF THE RESULTANT VECTOR • 2.PLACES THE RESULTANT IN EQUILIBRIUM (NO NET FORCE)

  2. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS • CREATING EQUILIBRIUM • A. “SIGN” EXAMPLE: • A 168 N sign is supported in a motionless position by two ropes that each make 22.5o angles with the horizontal. What is the tension in the ropes?

  3. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS • MOTION ALONG AN INCLINED PLANE A. INCLINED PLANE 1. Fg => the force “straight down” due to gravity . . . Fg = m g 2. FN => the force perpendicular to the surface the object is sitting on … perpendicular to the inclined plane (FN DOES NOT = Fg on an incline plane, FN = Fgy) 3. Ff = u FN. . . Ff => force “parallel” to the incline plane that resists movement “down” the incline plane 4. Fappl => force being applied parallel to the incline on the object to “force” it to overcome the Ff and begin accelerating (moving) down the incline 5. Fnet = m a => Fnet is the “resultant” force when Fappl is greater than the Ff

  4. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS • MOTION ALONG AN INCLINED PLANE A. INCLINED PLANE INCLINED PLANE EXAMPLE: A trunk weighing 562 N is resting on a plane inclined 30o above the horizontal. Find the components of the weight force parallel (Fgx) and perpendicular (Fgy) to the plane.

  5. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS • MOTION ALONG AN INCLINED PLANE A. INCLINED PLANE 6. Fnet = Fappl – Ff a. Fnet = m a . . . “resultant force” down the incline b. Fappl = Fgx. . . the force being applied parallel to the incline by an “agent” - Fgx is the “x vector component” of Fg which represents the hypotenuse of the vector triangle created by the object sitting on the incline plane - since sine (angle of inclination) = Fgx / Fg . . . Fgx = sine (angle of inclination) Fg

  6. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS • MOTION ALONG AN INCLINED PLANE A. INCLINED PLANE 7. Fnet = Fappl – Ff => m a = sine (angle of inclination) Fg – Ff a. Ff = u FN . . . m a = sine (angle of incl) Fg – u FN b.since FN is force perpendicular to the incline, FN = Fgy c. since cosine (angle of inclination) = Fgy / Fg . . . Fgy = cosine (angle of inclination) Fg 8. m a = sine (angle of inclination) Fg – Ff a.m a = sine (aoi) Fg – u Fgy b.m a = sine (aoi) Fg – u cosine (aoi) Fg 9. Since Fg = m g . . . m a = sine (aoi) m g – u cosine (aoi) m g a. m a = m g [sine (aoi) – u cosine (aoi)] b.a = g [sine (aoi) – u cosine (aoi)]

  7. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS • MOTION ALONG AN INCLINED PLANE B. “SKIING” DOWNHILL EXAMPLE: A 62 kg person on skis is going down a hill sloped at 37o. The coefficient of kinetic friction between the skis and the snow is 0.15. How fast is the skier going 5.0 s after starting from rest? hint: vf = vi + a t

  8. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS IV. PROJECTILEMOTION (video demonstration: Monkey & Hunter) A. HORIZONTAL PROJECTILE MOTION 1. TIME IS THE KEY COMPONENT! - THE TIME IT TAKES FOR A HORIZONTAL PROJECTILE TO HIT THE GROUND IS EQUAL TO THE TIME IT WOULD TAKE FOR THE SAME OBJECT TO HIT THE GROUND IF IT WERE SIMPLY DROPPED! (tvert = thoriz) - dvert = ½ g t2 - dhoriz = di + vhoriztT . . . diusually equals 0, vhoriz = vi

  9. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS IV. PROJECTILE MOTION • A. HORIZONTAL PROJECTILE MOTION • Example Problem: • A stone is thrown horizontally at 15 m/s in the direction of a • ginger. The teacher is standing on a cliff 44 m high. • If the ginger is on the ground below, how far from the base of the cliff should the said teacher lure the ginger into standing so said ginger can catch said stone? • How fast is said stone moving prior to smoking said ginger? • Hint: vf is the resultant of vvert and vhoriz

  10. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS V. PROJECTILES “LAUNCHED” AT AN ANGLE A. “GENERIC” ISSUES FOR PROJECTILE PROBLEMS 1. tup = tdown 2. tup + tdown = tT 3. dvert => how high did the projectile rise? 4. dhoriz => RANGE . . . How far did the projectile travel horizontally? 5. Distinguish between vhoriz, vvert and vresultant / vprojectile 6. vhoriz => vx 7. vvert => vy 8. To find maximum height . . . dh = vytup – ½ g tup2 9. To find tup or tdown . . . vy = a t . . . vy2 = 2 g dup/down 10. To find RANGE . . . dhoriz = vxtT 11. To find the vi / vresult / vproj … dT = di + vi t – ½ g t2

  11. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS V. PROJECTILES “LAUNCHED” AT AN ANGLE B. EXAMPLE PROJECTILE PROBLEM A ball is launched with an initial velocity of 4.47 m/s at an angle of 66oabove the horizontal. 1. How long does it take the ball to return to the launching height? 2. What was the maximum height the ball attained? 3. What was the range of the ball? 4. If someone named “Nate” or “Cam” was standing at the max range and the ball is was somewhat football shaped would he still drop it??? Just asking . . .

  12. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS V. PROJECTILES “LAUNCHED” AT AN ANGLE B. EXAMPLE PROJECTILE PROBLEM An arrow is shot at a target that is 60 m away. The arrow is projected at an angle of 30o, and is released at the same height as the target. 1) What was the initial velocity of the arrow? HINT: Without height, time, or any velocity components given, you must use the Total Displacement formula (Total Distance). dT = di + vyt – ½ g t2 - since dT = 0 m (arrow returns to same height), and di = 0 m - 0 = 0 + vyt – ½ g t2 (where vy = vertical component of velocity) - vi (initial velocity) is the hypotenuse of the vector triangle

  13. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS V. PROJECTILES “LAUNCHED” AT AN ANGLE B. EXAMPLE PROJECTILE PROBLEM dT = di + vyt – ½ g t2 - since dT = 0 m (arrow returns to same height), and di = 0 m - 0 = 0 + vyt – ½ g t2 (where vy = vertical component of velocity) - 0 = vyt – ½ g t2 . . . 0 = (vy – ½ g t) t - THEREFORE . . . t = 0 OR vy – ½ g t = 0 - vi (initial velocity) is the hypotenuse of the vector triangle - dx(range) = vx t . . . Cos A = vx / vi . . . vx = vi (cos 0) - dx = vi (cos A) t . . . t = dx / vi (cos A) - sine A = vy / vi . . . vy = vi (sine A)

  14. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS V. PROJECTILES “LAUNCHED” AT AN ANGLE B. EXAMPLE PROJECTILE PROBLEM dT = di + vyt – ½ g t2 - vy – ½ g t = 0 - vv is the hypotenuse of the vector triangle - dx(range) = vx t . . . Cos 0 = vx / vi . . . vx = vi (cos 0) - dx = vi (cos 0) t . . . t = dx / vi (cos 0) - sine 0 = vy / vi . . . vy = vi (sine 0) - vy – ½ g t = 0 . . . vi (sine 0) – ½ g ( dx / vi (cos 0) = 0 - vi (sine 0) = ½ g ( dx / vi (cos 0) - vi (sine 0) [ vi (cos 0) ] = ½ g dx

  15. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS V. PROJECTILES “LAUNCHED” AT AN ANGLE B. EXAMPLE PROJECTILE PROBLEM dT = di + vyt – ½ g t2 - vi (sine 0) [ vi (cos 0) ] = ½ g dx - vi2 (sine A) (cos A) = ½ g dx - vi2 = _____g dx______ 2 (sine A) (cos A) T H E R E F O R E : - vi2 = ____9.8 m/s2 (60 m)__ 2 (sine 30o) (cos 30o) - vi = 26.1 m/s

  16. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS CIRCULAR MOTION (ANGULAR MOTION) A. The distance around the perimeter of a circle is equal to the “angular” velocity times the time . . . 1. d = v t 2. d(for a circle) = 2 Π r 3. 2 Π r = v t . . . t around a circle is T (period) 4. v = 2 Π r T 5. Centripetal acceleration . . . ac = v2 r 6. ac = (2 Π r / T)2 r 7. ac = 4 Π2 r2 / T2 r 8. ac = 4 Π2 r T2

  17. PHYSICS CHAP 7: FORCES IN TWO DIMENSIONS CIRCULAR MOTION (ANGULAR MOTION) B. F = m ac 1. F = m v2 r 2. F = m 4 Π2 r T2

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