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Chapter 4. Motion With a Changing Velocity. P2.27: Find the magnitude and direction of the vector with the following components: x = -5.0 cm, y = +8.0 cm F x = +120 N, F y = -60.0 N v x = -13.7 m/s, v y = -8.8 m/s a x = 2.3 m/s 2 , a y = 6.5 cm/s 2.
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Chapter 4 Motion With a Changing Velocity
P2.27: Find the magnitude and direction of the vector with the following components: • x = -5.0 cm, y = +8.0 cm • Fx = +120 N, Fy = -60.0 N • vx = -13.7 m/s, vy = -8.8 m/s • ax = 2.3 m/s2, ay = 6.5 cm/s2
P2.56: A 3.0 kg block is at rest on a horizontal floor. If you push horizontally on the block with a force of 12.0 N, it just starts to move. • What is the coefficient of static friction? (b) A 7.0-kg block is stacked on top of the 3.0-kg block. What is the magnitude F of the force acting horizontally on the 3.0-kg block as before, that is required to make the two blocks start to move?
P3.47: A 2010-kg elevator moves with an upward acceleration of 1.50 m/s2. What is the tension that supports the elevator? P3.48: A 2010-kg elevator moves with a downward acceleration of 1.50 m/s2. What is the tension that supports the elevator?
Kinematic Equations for Const. Acceleration Fnet = ma. If Fnet is const, a will also be const. Uniformly accelerated motion: a = const.
A car moves at a constant acceleration of magnitude 5 m/s2. At time t = 0, the magnitude of its velocity is 8 m/s. What is the magnitude of its velocity at (i) t = 2s? (ii) t = 4s? (iii) t = 10s? A car moves at a constant acceleration of magnitude 5.7 m/s2. At time t = 0, the magnitude of its velocity is 18.3 m/s. What is the magnitude of its velocity at t = 2.2s?
Kinematic Equations for Const. Acceleration • Consider an object on which a net force Fnet acts on it. Thus it moves with an acceleration. • As the object moves, its velocity changes. Fnet Fnet Fnet a a Time = 0 Initial position = x0 Initial velocity = v0 Time = t Final position = x Final velocity = v
Kinematic Equations for Const. Acceleration Fnet = ma. If Fnet is const, a will also be const. Uniformly accelerated motion: a = const. aave = ainst Let us use initial time, t1 = 0. Final time, t2 = t, hence t = t – 0 = t Position: initial, x1 = x0, final, x2 = x Velocity, initial v1 = v0, final, v2 = v
Kinematic Equations for Constant Acceleration Uniformly accelerated motion: a = constant. Time: Initial = 0, final = t Positions: Initial = x0, final =x Velocity: Initial = v0, final = v v = v0 + at x = x0 +vot + ½ at2 v2 = v02 + 2a(x-x0) Average velocity vav = (v0 + v)/2
Example Problem 4.14 A train traveling at a constant speed of 22 m/s, comes to an incline with a constant slope. While going up the incline the train slows down with a constant acceleration of magnitude 1.4 m/s2. • Draw a graph of vx versus t. • What is the speed of the train after 8.0s on the incline? • How far has the train traveled up the incline after 8.0 s?
A car moving south slows down with at a constant acceleration of 3.0 m/s2. At t = 0, its velocity is 26 m/s. What is its velocity at t = 3 s? • 35 m/s south • 17 m/s south • 23 m/s south • 29 m/s south • 17 m/s north
A car initially traveling at 18.6 m/s begins to slow down with a uniform acceleration of 3.00 m/s2. How long will it take to come to a stop? • 55.8 s • 15.6 s • 6.20 s • 221.6 s • None of these
Free Fall • Free fall: Only force of gravity acting on an object making it fall. • Effect of air resistance is assumed negligible. • Force of gravity acting on an object near the surface of the earth is F = W = mg. • Acceleration of any object in free fall: a = g = 9.8 m/s2 down (ay = -9.8 m/s2).
Free Fall ay = -9.8 m/s2 ax = 0 +y +x
Free Fall contd… 1. a = g, regardless of mass of object.
2. a = g, regardless of initial velocity ay = -9.8 m/s2,ax = 0 +y v0 = 0 v0 = -15 m/s v0 = +15 m/s +x
3. Free Fall: Motion is symmetric. ay = -9.8 m/s2,ax = 0 At the maximum height: • vy = 0 • Speed at equal heights will be equal. • Equal time going up and down. +y v0 = +5 m/s +x
Example: Problem 4.32 A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s and the stone is 1.5 m above the ground when launched. • How high above the ground does the stone rise? • How much time elapses before the stone hits the ground?
APPARENT WEIGHT N mg A physics student whose mass is 40 kg stands inside an elevator on a scale that reads his weight in Newtons. Scale Reading = Normal force the scale exerts on the student. Scale Reading = N = mg = 40 x 9.8 N = 392 N
1. Elevator at rest. What will be the scale reading? N Fnet = N – W = may At rest means ay = 0. Hence N = W, ie apparent weight = true weight = 40 x 9.8 = 392 N W = mg
2. Elevator accelerating upwards with ay = 2.0 m/s2. What will be the scale reading?
N a W = mg 2. Elevator accelerating upwards with ay = 2.0 m/s2. What will be the scale reading? Fnet = N – W = may ay = + 2.0 m/s2 (positive because acceleration is upwards) . Hence, N –W = N – mg = may. N = mg + may = m(g+ay) = 40(9.8 + 2.0) = 472 N ie, apparent weight is greater than the true weight.
3. Elevator accelerating downwards with ay = 2.0 m/s2. What will be the scale reading?
N a W = mg 3. Elevator accelerating downwards with ay = 2.0 m/s2. What will be the scale reading? Fnet = N – W = may ay = - 2.0 m/s2 (negative because acceleration is downwards) . Hence N –W = N – mg = -may. N = mg - may = m(g - ay) = 40(9.8 - 2.0) = 312 N ie, apparent weight is less than the true weight.
A 112.0-kg person stands on a scale inside an elevator moving downward with an acceleration of 1.80 m/s2. What will be the scale reading? • 1299 N • 1,098 N • 896 N • 112 N • 0 N
A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise? • 9.8 m • 3.00 m • 1.15 m • 26.1 m • 19.6 m
WEIGHTLESSNESS If the elevator was going down with an acceleration ay = g = -9.8m/s2, then N = m(g-g) = 0 ie, apparent weight = 0 This is “weightlessness” or “zero gravity” Apparent weight of an object in free fall is zero while its true weight remains unchanged.
Equilibrium Newton’s 2nd Law: Fnet = ma For an object in equilibrium: Fnet = 0 Static (v = 0) 0r dynamic (v = constant) eqlbm. 2-dimensions, separate the x and y components and treat the problem as two 1-dim problems. Fx = max Fy = may For equilibrium,Fx = max = 0 and Fy = may = 0
2-Dimensions y x • X and Y are INDEPENDENT! • Break 2-D problem into two 1-D problems.
Equilibrium Determine the tension in the 6 m rope if it sags 0.12 m in the center when a gymnast with weight 250 N is standing on it. y x direction: Fx = max = 0 -TL cosq + TR cosq = 0 TL = TR TR TL x W TR y direction: Fy = may = 0 TL sinq + TR sinq - W = 0 2 T sinq = W T = W/(2 sinq) = 3115 N .12 m q 3 m
N N fk fs F F W W Equilibrium on a Horizontal Plane Object at rest or moving with const. velocity • Fx = max = 0 and Fy = may = 0 Object at rest Sliding with constant velocity No motion until F = > fsmax Fx = F - fs = 0 or F = fs. Fy = N - W = 0 or N = W Fx = F - fk = 0 or F = fk Fy = N - W = 0 or N = W
Equilibrium on an inclined Plane y fs y x fs N x W.cos N W Wy W.sin W Wx y W fs x N W.sin W.cos An object at rest on an inclined plane • Fx = max = 0 and Fy = may = 0
Equilibrium on an inclined Plane y fs x N W.sin W.cos An object at rest on an inclined plane Fx = max = 0 and Fy = may = 0 Fy = may = 0 N - Wcos = 0 or N = Wcos Fx = max = 0 fs - Wsin = 0 or fs = Wsin If angle is increased, the object will eventually slide down the plane. Sliding will start beyond angle max At max: fsmax = W.sinmax. But fsmax = sN = s(Wcosmax) Therefore, sWcosmax = Wsin max ORs = (Wsinmax)/Wcosmax ie, s = tanmax
Equilibrium on an inclined Plane y fs x N W.sin W.cos An object at rest on an inclined plane N = Wcos fs = Wsin • If angle is increased, the object will eventually slide down the plane. • Sliding will start beyond angle max • At max: fsmax = W.sinmax. • But fsmax = sN = s(Wcosmax) • Therefore, sWcosmax = Wsin max • ORs = (Wsinmax)/Wcosmax ie, s = tanmax
A mass m being pulled uphill by a force F y F x If m = 510 kg, s= 0.42, k = 0.33, = 15o: • Find minimum force F needed to start the mass moving up. • If the force in (a) is maintained on the mass, what will its acceleration be? N W.sin fk W.cos (c) To move the mass with constant speed, what must the value of F be?
A block is at rest on a flat board. The flat board is gently tilted. At what angle will the block start to slide? Assume the coefficient of static friction (s) between the block and the board is 0.48. • 0.48o • 61.3o • 28.7o • 25.6o • 0.00837o
Position, Velocity and Acceleration • Position, Velocity and Acceleration are Vectors! • x and y directions are INDEPENDENT! y direction x direction
Velocity in Two Dimensions A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the change in velocity? x-direction vix = 5 m/s vfx = 3 m/s cos(25) Dvx = 3cos(25)–5 =-2.28m/s y-direction viy = 0 m/s vfy = 3 m/s sin(25) Dvy = 3sin(25)=+1.27 m/s y x 3 m/s 5 m/s
Acceleration in Two Dimensions A ball is rolling on a horizontal surface at 5 m/s. It then rolls up a ramp at a 25 degree angle. After 0.5 seconds, the ball has slowed to 3 m/s. What is the average acceleration? [Assume force of gravity is very small]. y x-direction y-direction x 3 m/s 5 m/s
A wagon of mass 50 kg is being pulled by a force F of magnitude 100 N applied through the handle at 30o from the horizontal. Ignoring friction, find the magnitude of • the horizontal component of F. • the horizontal component of acceleration. • the normal force exerted onthe wagon.
Projectile Motion A projectile – An object moving in 2-dimensions near the surface of the earth with only the force of gravity acting on it. Eg: golf ball, batted base ball, kicked football, soccer ball, bullet, etc. • Assume no air resistance. • Assume g = -9.8 m/s2 constant. • We are not concerned with the process that started the motion!
Free Fall: 1-dimensional motion. ay = -9.8 m/s2,ax = 0 +y At the maximum height: • vy = 0 • Speeds at equal heights will be equal. • Equal time going up/down. v0 = +5 m/s +x
PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2, ax = 0 +y v0 = 5 m/s v0y +x v0x
PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2 ax = 0 v0x = v0cos v0y = v0sin What will happen to the y-component of the velocity? What will happen to the x-component of the velocity?
Kinematics in Two Dimensions • x = x0 + v0xt + ½ axt2 • vx = v0x +axt • vx2 = v0x2 + 2ax(x - x0) • y = y0 + v0yt + ½ ayt2 • vy = v0y +ayt • vy2 = v0y2 + 2ay(y – y0) x and y motions are independent! They share a common time t.
Kinematics for Projectile Motionax = 0 ay = -g • y = y0 + v0yt - 1/2 gt2 • vy = v0y-gt • vy2 = v0y2 - 2g y • x = x0 + v0t • vx = v0x X Y
PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2,ax = 0 Once the projectile is in air, the only force acting on it is gravity. Its trajectory (path of motion) is a parabola. Fnet = ma = -mg ay = -9.8 m/s2 ax = 0
PROJECTILE: Free Fall motion in 2-dimensions. ay = -9.8 m/s2 andax = 0 v0x = v0cos and v0y = v0sin
Two balls A and B of equal mass m. Ball A is released to fall straight down from a height h. Ball B is thrown horizontally. Which ball lands first? A B h ay = -9.8 m/s2 ax = 0 Vo = 0 v0x = 0 v0y = 0 ay = -9.8 m/s2 ax = 0 Vo 0 v0x = Vo v0y = 0
A • y = y0 + v0yt + ½ ayt2 • vy = v0y +ayt • vy2 = v0y2 + 2ay(y – y0) To find time t, use y = y0 + v0yt + ½ ayt2 -h = 0 + (0 . t) + ½ (-g)t2 Gives 2h = gt2 and t = (2h/g) ay = -9.8 m/s2 ax = 0 vo = 0 v0x = 0 v0y = 0 y0 = 0, y = -h