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Lecture 7 Goals: Identify the types of forces Use a Free Body Diagram to solve 1D and 2D problems with forces in equilibrium and non-equilibrium (i.e., acceleration) using Newton’ 1st and 2nd laws. Distinguish static and kinetic coefficients of friction 1st Exam Thursday, Oct. 6 from 7:15-8:45 PM Chapters 1-7(light) Physics 207: Lecture 7, Pg 1 Chaps. 5, 6 & 7 What causes motion? (actually changes in motion) What kinds of forces are there ? How are forces & changes in motion related? Physics 207: Lecture 7, Pg 2 Newton’s First Law and IRFs An object subject to no external forces moves with constant velocity if viewed from an inertial reference frame (IRF). If no net force acting on an object, there is no acceleration. The above statement can be used to define inertial reference frames. Physics 207: Lecture 7, Pg 3 IRFs An IRF is a reference frame that is not accelerating (or rotating) with respect to the “fixed stars”. If one IRF exists, infinitely many exist since they are related by any arbitrary constant velocity vector! In many cases (i.e., Chapters 5, 6 & 7) the surface of the Earth may be viewed as an IRF Physics 207: Lecture 7, Pg 4 No net force No acceleration F Fnet 0 a 0 If zero velocity then “static equilibrium” If non-zero velocity then “dynamic equilibrium” Forces are vectors F Fnet F1 F2 F3 Physics 207: Lecture 7, Pg 5 Newton’s Second Law The acceleration of an object is directly proportional to the net force acting upon it. The constant of proportionality is the mass. This is a vector expression: Fx, Fy, Fz Units The metric unit of force is kg m/s2 = Newtons (N) The English unit of force is Pounds (lb) Physics 207: Lecture 7, Pg 6 Example Non-contact Forces All objects having mass exhibit a mutually attractive force (i.e., gravity) that is distance dependent At the Earth’s surface this variation is small so little “g” (the associated acceleration) is typically set to 9.80 or 10. m/s2 FB,G Physics 207: Lecture 7, Pg 7 Contact (e.g., “normal”) Forces Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point). Here: A contact force from the table opposes gravity, Normal force is perpendicular to the surface. FB,T Physics 207: Lecture 7, Pg 8 No net force No acceleration F Fnet ma 0 Fx 0 Fy 0 y (Force vectors are not always drawn at contact points) FB,T Normal force is always to a surface Fy mg N 0 FB,G N mg Physics 207: Lecture 7, Pg 9 High Tension A crane is lowering a load of bricks on a pallet. A plot of the position vs. time is There are no frictional forces Compare the tension in the crane’s wire (T) at the point it contacts the pallet to the weight (W) of the load (bricks + pallet) Time Height A: T > W B: T = W C: T< W D: don’t know Physics 207: Lecture 7, Pg 10 Important notes Many contact forces are conditional and, more importantly, they are not necessarily constant We have general idea of forces from everyday life. In physics the definition must be precise. A force is an action which causes a body to accelerate. (Newton’s Second Law) On a microscopic level, all forces are non-contact Physics 207: Lecture 7, Pg 11 Analyzing Forces: Free Body Diagram A heavy sign is hung between two poles by a rope at each corner extending to the poles. Eat at Bucky’s A hanging sign is an example of static equilibrium (depends on observer) What are the forces on the sign and how are they related if the sign is stationary (or moving with constant velocity) in an inertial reference frame ? Physics 207: Lecture 7, Pg 12 Free Body Diagram Step one: Define the system T2 T1 q2 q1 Eat at Bucky’s mg T2 T1 q1 q2 mg Step two: Sketch in force vectors Step three: Apply Newton’s 2nd Law (Resolve vectors into appropriate components) Physics 207: Lecture 7, Pg 13 Free Body Diagram T1 T2 q2 q1 Eat at Bucky’s mg Vertical : y-direction Horizontal : x-direction 0 = -mg + T1 sinq1 + T2 sinq2 0 = -T1 cosq1 + T2 cosq2 Physics 207: Lecture 7, Pg 14 Scale Problem You are given a 5.0 kg mass and you hang it directly on a fish scale and it reads 50 N (g is 10 m/s2). 50 N 5.0 kg Now you use this mass in a second experiment in which the 5.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale. What does the string do? What does the pulley do? What does the scale now read? ? 5.0 kg Physics 207: Lecture 7, Pg 15 Pushing and Pulling Forces String or ropes are examples of things that can pull You arm is an example of an object that can push or push Physics 207: Lecture 7, Pg 16 Examples of Contact Forces: A spring can push Physics 207: Lecture 7, Pg 17 A spring can pull Physics 207: Lecture 7, Pg 18 Ropes provide tension (a pull) In physics we often use a “massless” rope with opposing tensions of equal magnitude Physics 207: Lecture 7, Pg 19 Moving forces around Massless strings: Translate forces and reverse their direction but do not change their magnitude (we really need Newton’s 3rd of action/reaction to justify) string T1 -T1 Massless, frictionless pulleys: Reorient force direction but do not change their magnitude T2 T1 -T1 | T1 | = | -T1 | = | T2 | = | T2 | -T2 Physics 207: Lecture 7, Pg 20 Scale Problem You are given a 5.0 kg mass and you hang it directly on a fish scale and it reads 50 N (g is 10 m/s2). 50 N 5.0 kg Now you use this mass in a second experiment in which the 5.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale. What force does the fish scale now read? ? 5.0 kg Physics 207: Lecture 7, Pg 21 What will the scale read? A 25 N B 50 N C 75 N D 100 N E something else Physics 207: Lecture 7, Pg 22 Scale Problem Step 1: Identify the system(s). In this case it is probably best to treat each object as a distinct element and draw three force body diagrams. One around the scale One around the massless pulley (even though massless we can treat is as an “object”) One around the hanging mass Step 2: Draw the three FBGs. (Because this is a now a one-dimensional problem we need only consider forces in the y-direction.) ? 5.0 kg Physics 207: Lecture 7, Pg 23 Scale Problem 3: T” T’ 1: 2: T ? W -T ’ S Fy = 0 in all cases 1.0 kg -T -T ? -mg 1: 0 = -2T + T ’ 2: 0 = T – mg T = mg 3: 0 = T” – W – T ’ (not useful here) Substituting 2 into 1 yields T ’ = 2mg = 100 N (We start with 50 N but end with 100 N) 5.0 kg Physics 207: Lecture 7, Pg 24 Home Exercise, Newton’s 2nd Law A woman is straining to lift a large crate, without success. It is too heavy. We denote the forces on the crate as follows: P is the upward force being exerted on the crate by the person C is the contact or normal force on the crate by the floor, and W is the weight (force of the earth on the crate). Which of following relationships between these forces is true, while the person is trying unsuccessfully to lift the crate? (Note: force up is positive & down is negative) A. B. C. D. P+C<W P+C>W P=C P+C=W Physics 207: Lecture 7, Pg 25 A “special” contact force: Friction What does it do? It opposes motion (velocity, actual or that which would occur if friction were absent!) How do we characterize this in terms we have learned? Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then “inferred”)! j N FAPPLIED ma fFRICTION i mg Physics 207: Lecture 7, Pg 26 If no acceleration No net force So frictional force just equals applied force Key point: It is conditional! N FAPPLIED fFRICTION j i mg Physics 207: Lecture 7, Pg 27 Friction... Friction is caused by the “microscopic” interactions between the two surfaces: Physics 207: Lecture 7, Pg 28 Friction: Static friction Static equilibrium: A block with a horizontal force F applied, S Fx = 0 = -F + fs fs = F FBD S Fy = 0 = - N + mg N = mg As F increases so does fs N F m fs 1 mg Physics 207: Lecture 7, Pg 29 Recap Assignment: Soon….HW4 (Chapters 6 & 7, due 10/4) Read through first half of Chapter 7 Physics 207: Lecture 7, Pg 30