Dynamics the branch of mechanics that deals with the motion and equilibrium of systems under the action of forces, usually from outside the system. 40S Physics Cara Katz, Yael Shrom, Hannah Glinter

Outline of Unit:  30S Review of Forces ~ formulas ~ formulas ~ Net Force ~ Net Force ~ Free Body Diagrams ~ Free Body Diagrams ~ Forces and Motions (friction) ~ Forces and Motions (friction)  Forces in 2-D  Equilibrium in 1-D and 2-D  Forces at an angle  Inclined Planes

30S Review:  Formulas: ~ Fg=mg ~ Fg=mg ~ Fnet=ma ~ Fnet=ma ~ Ff=µFn ~ Ff=µFn Example Question: A box has a mass of 5.0 kg. The applied force is 25 N. µk=0.2. Find Fg, Fn, Ff and the acceleration of the box. Step 1: Fg=mg, Fg= (5.0)(-9.8), Fg= 49 N (down) Step 2: Fn=Fg, Fn= 49 N (up) Step 3: Ff=µFn, Ff= (0.2)(49), Ff= 9.8 N (left) Step 4: Fnet=ma, 15.2 N/5.0= a, a= 3.04 m/s ²

30S Review: NNNNet Force ~ vector sum of all the forces which act upon an object (sum of all the forces)

30S Review:  Free Body Diagrams

30S Review:  Forces and Motion (Friction) ~ using the formulas: ~ using the formulas:Example A force of 60 N is applied to a 20 kg. mass on a surface, where the force of the friction is 25 N. At what rate does this mass accelerate? A force of 60 N is applied to a 20 kg. mass on a surface, where the force of the friction is 25 N. At what rate does this mass accelerate? Fa= 60N m= 20 kg. Ff= 25 N Fnet= 60-25= 35 N a= Fnet/m, a=35/20, Fnet= 1.75m/s²

Forces in 2-D  Things to remember: ~ change all diagonal forces into x and y components ~ change all diagonal forces into x and y components ~ analyze y and x axis SEPARATELY! ~ analyze y and x axis SEPARATELY! ~ solve for Fnet x, Fnety and Fnet ~ solve for Fnet x, Fnety and Fnet ~ solve for Fn last ~ solve for Fn last Example Questions Example Questions  Case 1: object on surface  Case 2: Applying a force to the object  Case 3: Pulling at an angle  Case 4: Pushing at an angle

Forces in 2-D  Case 1: Object sitting on a surface Y axis Fg= 49 N down Fn= 49 N up Fnety= 0 N X axis No forces Fnetx= 0N Fnet=0 **** Fnet= Fnetx and Fnety=0 as long as object is not falling!!!! Case 2: Applying a Force Y axis Fg= 196 N (down) Fn= 196 N (up) Fnety= 0N X axis Ff= 10N left Fa= 60 N right Fnetx= 50 N right Fnet= 50 N right

Forces in 2-D  Case 3: Pulling at an angle Y axis Fn = 59N (up) Fg = 49N (down) FAY = 10N (up) Fnet Y = 0N X axis Ff = 20N (left) FAX = 30N (right) Fnet X = 10N (right) Fnet = 10N (right) ****** If you were to look for acceleration, you use Fnet = ma In this case, 10 = 5a m/s² Acceleration = 2m/s²

Forces in 2-D  Case 4: Pushing at an Angle Y Axis Fn = 69N (up) Fg = 49N (down) FAY = 20N (down) Fnet Y = 0N X Axis Ff = 20N (left) FAX = 50N (right) Fnet X = 30N (right) Fnet = 30N (right) **** Using what you know, find the acceleration!

Equilibrium AAAAn Object is in equilibrium when it has zero acceleration. FFFFnet must be zero in x and y directions 1111-D Problems: Word Problem Suppose you had a sign with a mass of 20.0 kg. suspended by two ropes, one on each side of the sign. What would be the tension in each of the ropes? Answer: Fg= mg Fg= (20)(9.8) Fg= 196 N (down) Therefore each rope has a tension of 98 N. Picture Problem What is the Fnet and is this an equilibrium? Fnet= 5 N (left)=> therefore because there is an Fnet that is not 0, this is not an equilibrium

Equilibrium  2-D Problems  Example Problem Suppose a sign is hanging from 2 wires, as shown in the diagram. The sign in 30 kg. What is the tension is each of the ropes? Suppose a sign is hanging from 2 wires, as shown in the diagram. The sign in 30 kg. What is the tension is each of the ropes? T3= Fg Fg= (30)(9.8) Fg= 294 N (down) Sin45= 147/ T1 T1= N Sin37= 147/ T2 T2= N

Forces at an Angle  It requires less force to pull than push (this is due to the decreased normal force, which decreases the force of kinetic friction.  Hint: this is the same as Case 3 and 4, but this time you’re given the angle!  Example Problem A 20.0 Kg box is dragged along a horizontal floor with a force of N. The force is applied at 40.0 ° above the horizontal and to the right. If the coefficient of kinetic friction is 0.30, what is the net force on the box, and what is its acceleration? Fg= 196 N Cos40=Fax/100, Fax=76.6 N (right) Sin40=Fay/100, Fay= 64.3 N (up) Fn=Fg-Fay, Fn= N Ff= 39.5 N Fnet= 37.1 N (right) (Fax- Ff) m/s² a= 37.1/20, a= 1.86 m/s² *** Using the kinematics equations, you can find time is asked to! Remember v1=0

Inclined Planes Things To Remember: 1) Fn= Fgy 2) Fnet= Fgx (because there is no Ff) 3) Fgx= (Fg)(sin θ) ~ Fgx- also known as “parallel to plane” 4) Fgy= (Fg)(cos θ) ~ Fgy- also known as perpendicular force