1. Unbalanced Forces

2. A Review of Newton’s 1 st Law: “An object in motion will tend to stay in motion & an object at rest will tend to stay at rest unless acted upon by an outside (unbalanced) Force.” IN OTHER WORDS: –Already in motion? Keep going! –Not moving? Stay where you are!

3. A Review of Newton’s 2 nd Law: Force = mass * acceleration (F=m*a) –Also… F net =m*a net “net” means “overall” So we can look at Forces both individually and all acting together, depending on our needs.

4. A Review of Newton’s 3 rd Law: “For every action there is an equal and opposite reaction.” –…as when the Normal Force (N) acting on an object is equal to the component of the Force in the opposite direction: FgFg F g,x F g,y

5. A Review of Newton’s 3 rd Law: Since there is no acceleration in the y- direction, N=-F g,y. FgFg F g,x F g,y N If we assume that the block is not sliding down the ramp, f s must be f s =-F g,x. fsfs

6. Discussion: When the application of Force results in motion, there are two possible scenarios: CASE 1: Constant Speed F g =-F air “terminal velocity” FgFg F air net Force = zero! So net acceleration = zero! CASE 2: Changing Speed F kick FgFg starting from rest projectile motion! net Force is NOT zero! So net acceleration is NOT zero!

7. Example #1: 15 kg UPS box pushed across a floor: CONSTANT FORCE APPLIED of F push = 500 Newtons μ k = 0.62 (f k = μ k *N) F push fkfk Don’t forget the vertically oriented Forces! FgFg N What is the net Force, net acceleration, and how far has the box been pushed after 30 seconds? TRY IT YOURSELF, FIRST!

8. Example #1: F net =? a net =? Net vertical Force is zero (so a net,vertical = 0 m/s 2 ) and we know it is not moving in the vertical direction. That was easy. Horizontally: F net = F push + f k F push fkfk FgFg N F net = 500 N + f k (let friction be neg.) F net = 500 N + -(0.62*N) F net = 500 N + -[0.62*(15kg*9.8m/s 2 )] F net = 500 N + -91.14 N = 408.86 N F net = 408.86 N, m = 15 kg, then a net = ?

9. Example #1: F net =? a net =? F push fkfk FgFg N F net = 408.86 N, m = 15 kg, then a net = ? Remember Newton’s 2 nd Law: F net =m*a net Using Algebra: a net = F net /m a net = 408.86N/15kg a net = 27.26 m/s 2

10. Example #1: Now, how FAR will it have gone?? F push fkfk FgFg N Remember, a net = 27.26 m/s 2 only is experienced horizontally in this case, so the package will only MOVE horizontally! We use this equation to find position after some time (30 s) when net- acceleration is nonzero: x t = x 0 + v 0 *t + ½*a net *t 2 It is already setup to find final position… x t = 0m + (0m/s*30s) + [½*27.26m/s 2 *(30s) 2 ] x t = ½*27.26m/s 2 *900s 2 x t = 12267 m !!! The “FLASH”

11. Example #2: Crate sliding down an inclined-plane: We set this one up just as we would any ramp problem: FgFg F g,x F g,y N fkfk N The Normal Force (N) is still the opposite of F g,y … The frictional force is now kinetic frictional force (f k ) and is less than F g,x … where μ k = 0.31 Since the x-component of the Weight (F g ) is not fully balanced by the kinetic frictional force (f k ), there will be a nonzero net-Force and therefore a nonzero net- acceleration! If the mass of the crate is 40 kg and Ө=30 ˚, how far down the ramp will the crate have traveled after 3 seconds? TRY IT YOURSELF, FIRST! Assume the crate was just placed there!

12. Example #2: Crate sliding down an inclined-plane: FgFg F g,x F g,y N fkfk N If the mass of the crate is 40 kg, how far down the ramp will the crate have traveled after 3 seconds? AS BEFORE: (no vertical motion) F net,x = F g,x + f k F net,x = F g *sin(Ө) + μ k *N (N = ?) N = -F g,y = -F g *cos(Ө) … (F g = ?) F g = m*g = (40kg*-9.8m/s 2 ) = -392 Newtons N = 392N*cos(30 ˚ ) = 339.48 Newtons Therefore…. F net,x = [-392N*sin(30 ˚ )] + [0.31*339.48N] = (-196 N) + (105.24 N) F net,x = -90.76 Newtons so… a net,x = (F net,x /m) = (-90.76N)/(40kg) = -2.27 m/s 2 (note negative!)

13. Example #2: Crate sliding down an inclined-plane: FgFg F g,x F g,y N fkfk N Therefore…. F net,x = [-392N*sin(30 ˚ )] + [0.31*339.48N] = (-196 N) + (105.24 N) F net,x = -90.76 Newtons so… a net,x = (F net,x /m) = (-90.76N)/(40kg) = -2.27 m/s 2 (note negative!) Since the crate started from rest once it was placed atop the ramp, x 0 and v 0 are both zero. (x t = x 0 + v 0 *t + ½*a*t 2 ) becomes… x t = ½*a*t 2 x t = (0.5)*(-2.27m/s 2 )*(3s) 2 x t = (-1.135 m/s 2 )*(9s 2 ) x t = -10.215 m (the crate has traveled 10.22 m DOWN THE RAMP)

14. Example #3: Car starts from rest: A constant net Force of 1000 N accelerates the car forward: Not being concerned with Rolling Friction or frictional forces acting within the mechanisms of the car, What must the mass of the vehicle be if the net- acceleration it experiences is 23 m/s 2 ? How fast is the car going after 6 seconds? How far has the car gone in that time? TRY IT YOURSELF, FIRST!

15. Example #3: Car starts from rest: What must the mass of the vehicle be if the net- acceleration it experiences is 23 m/s 2 ? How fast is the car going after 6 seconds? How far has the car gone in that time? F net = m*a net m = F net /a net m = 1000N/23m/s 2 v t = v 0 + a*t v t = (0 m/s) + (23 m/s 2 )*(6s) v t = 138 m/s x t = x 0 + v 0 *t + ½*a*t 2 x t = (0 m) + (0 m/s)*(6 s) + (0.5)*(23 m/s 2 )*(6s) 2 = 414 meters

16. Any Questions?