Chapter 4 Motion in tyo and three
Chapter 4 Motion in two and three dimensions
TWo principles for 2D and 3D motions 1)The principle of independence of force 2)The principle of superposition of motion F F3 F2
Two principles for 2D and 3D motions: 1) The principle of independence of force 2) The principle of superposition of motion F1 F2 F3
Scio4小 oijoririhree c9so5 wiin constant accelerator Now we consider a particle move in three dimensions with constant acceleration. We can represent the acceleration as a vector a=a ita i+ak The particle starts at t=0 with initial position -xo i+ yo j+=ok and an initial velocity vo=vo, i+vov j+vo k
The particle starts at t=0 with initial position and an initial velocity . Section 4-1 Motion in three dimensions with constant acceleration Now we consider a particle move in three dimensions with constant acceleration. We can represent the acceleration as a vector: → a = a i+ a j+ a k x y z → r = x i+ y j+ z k 0 0 0 o → v = v i+ v j+ v k 0 0x 0 y 0z
a,a constant L ar, a,a, all constants Vx vox to t +a t v=v+at(4-1) +ot In a similar way x=Xo+ vo.t y=Vo+volta,t r=r+vot+ at 2=z0+v=2+a2t (4-2)
In a similar way: v v a t x x x = + 0 v v a t y y y = + 0 v v a t z z z = + 0 v v a t → → → = 0 + a, a constant → , , , x y z a a a all constants x v x ax t x t 2 0 0 2 1 = + + z v z az t z t 2 0 0 2 1 = + + y v y ay t y t 2 0 0 2 1 = + + 2 0 0 2 1 r r v t a t → → → → = + + （4-1） （4-2）
co42小yon3 s lays intree clmersjona yector」for∫」 ∑ F=ma(43) Which includes the three component equations ∑∑∑ F=max F=mas F=ma,(4-4)
Section 4-2 Newton’s laws in three dimensional vector form (4-3) Which includes the three component equations (4-4) → → F = ma Fx = max Fy = may Fz = maz
Sample problem 1. A crate of mass m=o2 kg is sliding without friction with an initial velocity of v0=0.4 r/s along the floor. In an attempt to move it in a different direction, Torn pushes opposite to iis initial notion with a constant force of a nacinitude F=81N, while Jane pushes in al perpendicular direction with a constant force oi magnitude F2-105N If they each push for 3. Os. in what direction is the crate moving when they stop pushing?
Sample problem 1. A crate of mass m=62 kg is sliding without friction with an initial velocity of v0=6.4 m/s along the floor. In an attempt to move it in a different direction, Tom pushes opposite to its initial motion with a constant force of a magnitude F1=81N, while Jane pushes in a perpendicular direction with a constant force of magnitude F2=105N. If they each push for 3.0s, in what direction is the crate moving when they stop pushing?
Section 4-3 Projectile noor Figure 4-4 shows the initial motion of a projectile at the y instant of launch. Its initial velocity is vo, directed at an mg angle p from the horizontal. X We choose suitable coordinate Fig 4-4 A particle is system to make launched with initial 0 0 velocity
Section 4-3 Projectile motion Figure 4-4 shows the initial motion of a projectile at the instant of launch. Its initial velocity is , directed at an angle from the horizontal. o v → 0 0 x0 = 0 y0 = o v y → x 0 Fig 4-4 A particle is launched with initial velocity mg We choose suitable coordinate system to make: o
The components of the initial velocity are Po Voy =vo sin p(4-6) Gravity is the only force acting on the particle, so the components of the net force ae∑F=0∑ F 4-7 y 0 g 8) Ox 8(4-9) Position components. X=Vo Ox g(4-10) 2
The components of the initial velocity are (4-6) Gravity is the only force acting on the particle, so the components of the net force are (4-7) (4-8) (4-9) Position components: (4-10) 0 0 vox = v cos 0 0 voy = v sin Fx = 0 Fy = −mg ax = 0 ay = −g x x v v = 0 v v gt y = 0 y − x v t = 0x 2 0 2 1 y v t gt = y −
From Egs(4-10), we can eliminate t and obtain the relationship between x and y(after considering Eqs.(4-6) g - tan ox 2( Vo cos少) (4-13) which is the equation of a trajectory(轨线)of the projectile, the ecluation of a galraioolal Hence the trajectory of a projectile is parabolic
From Eqs. (4-10), we can eliminate t and obtain the relationship between x and y (after considering Eqs. (4-6)): 2 2 0 0 0 2( cos ) tan x v g y x = − (4-13) which is the equation of a trajectory (轨线) of the projectile, the equation of a parabola. Hence the trajectory of a projectile is parabolic
Fig 4-5 trajectory of a projectile y Ox vor X R
Fig 4-5 trajectory of a projectile v0x v0 y v0x vy R 0 v v0x v0x v0x v0 y x y o