Welcome to Physics 101! Lecture 01: Introduction to Forces

1. Welcome to Physics 101!Lecture 01: Introduction to Forces • Newton’s Laws • Forces • Free Body Diagrams

2. Newton’s 1st Law of Motion • If the sum of all external forces on an object is zero, then its speed and direction will not change. (Know as the Law of Inertia.) • Example: air hockey table • When the puck travels with no net force on it, it moves in a straight line with constant speed. • The only time the speed or direction changes is when a net force (from the walls or mallets) is exerted on the puck!

3. Newton’s 2nd Law of Motion • If a nonzero net force is applied to an object its motion will change. (The change is described by the equation F= ma.) • Example: dropping a book • When you drop a Calculus book from your 10th story dorm window, there is a net force acting on it (gravity) thus it begins to accelerate toward the street below!

4. Newton’s 3rd Law of Motion • In an interaction between two objects, the forces that each exerts on the other are equal in magnitude and opposite in direction. • Example: rowing a boat • When you row, you are pushing the water backwards… • Thus the water pushes your paddle (and therefore you) forward!

5. Forces • Quantifies “interactions” between objects. • 4 Fundamental Forces • Gravity • Electromagnetic • Strong Nuclear • Weak Nuclear • Forces are Vectors • Have magnitude and direction • SI units are Newtons (N)

6. Gravity • Any two objects with mass are attracted. • On the surface of the earth, the force of attraction is referred to the force of gravity or weight. • rearth = 6.4x106 m, mearth = 6x1024 kg, G = 6.67x10-11 N-m2/kg2 • Near the earth, the radius of the earth can be used for r. • Example: a person’s weight • Consider a person with a typical mass of 80 kg. • Near the earth, his weight (or force of gravity pulling him down toward the center of the earth) would be Fg = 784 N.

7. Contact Forces • An object in contact with a surface may have a normal force and force of friction acting on it. • The normal force is always perpendicular to the surface. There is no equation for the normal force – its magnitude depends on the situation. • The frictional force is always parallel to the surface (opposing motion). The frictional force depends on the “coefficient of friction” between the object and surface. • Kinetic friction (object sliding on a surface): Ff = k*FN • Static friction (no sliding): there is no definite equation for static friction, but its maximum value is given by: Ffs*FN • Example: a sled sliding across a snowy field The ground is pushing up on the sled (to keep it from falling into the earth), and the ground is pushing against the sled’s motion (slowing it down a little).

8. y x Free Body Diagrams • The key to success in Physics 101! • Simple picture of just the object of interest. • Choose coordinate system (x,y). • Identify all forces acting on object and draw arrows showing the directions. • Example: a dresser being pushed down the hall of your apartment as you move in. There are four forces: gravity, the normal force, friction, and you pushing on the dresser. FN Ff Fp Fg

9. Summary • Newton’s Laws of Motion • Inertia • F=ma • Equal and Opposite Force Pairs • Forces: • Non-Contact: Gravity • Contact: Friction and Normal • Free Body Diagrams • Isolate One Object • Each Direction is Independent

10. y x Example: The Dresser • Let’s return to the dresser FBD and do some calculations…(I will solve the horizontal direction and you will solve the vertical direction)… FN • Given: the dresser has a mass of 120 kg and you are pushing on it with a force of 345 N to make it slide down the hall at a constant velocity. Ff Fp Fg

11. y x Example: The Dresser m = 120 kg and Fp = 345 N • Once the FBD is drawn and we have picked a coordinate system we must apply Newton’s Second Law: FN F = ma We know the horizontal acceleration is zero (the velocity is constant). F = 0 Ff Fp Fg

12. y x Example: The Dresser m = 120 kg and Fp = 345 N • Now start to substitute in what is known: FN F = 0 We know there are two forces in the x-direction (the push and friction). Fp – Ff = 0 Ff Fp Fg

13. y x Example: The Dresser m = 120 kg and Fp = 345 N • Solve for the unknown: FN Fp – Ff = 0 The force of the push is given as 345 N. (345 N) – Ff = 0 Ff = 345 N Ff Fp Fg

14. y x Example: The Dresser • Follow-up: now we can find the coefficient of kinetic friction between the dresser and the floor… FN Ff = k*FN (345 N) = k*(1176 N) k = 0.29 Ff Fp Fg