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Physics 203 College Physics I Fall 2012. S. A. Yost. Chapter 4 . Newton’s Laws – Part 3. Weight and Normal Force. If you stand on a scale, what force does a scale measure? ( N or W ) It measures the normal force between you and the scale.
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Physics 203College Physics IFall 2012 S. A. Yost Chapter 4 Newton’s Laws – Part 3
Weight and Normal Force If you stand on a scale, what force does a scale measure? (N or W) It measures the normal force between you and the scale. This equals the weight by Newton’s second law – provided you are in an inertial reference frame. N W N+ W = 0
Weight in an Elevator a = 2.2 m/s2 An elevator accelerates upward at 2.2 m/s2. What does the passenger apparently weigh, as measured by the scale in the elevator, if his mass is 100 kg?
Weight in an Elevator a = 2.2 m/s2 A net force on the passenger causes the upward acceleration: F = N – W = ma. W = mg. N = ma + W = m(a+g) = (100 kg)(12.0 m/s2) = 1200 N. N W
Tension • Pulling on a rope applies tension. The force always acts along the rope if it is free to bend.
Measuring Tension • The magnitude of the tension in the rope shown is the magnitude of the force F pulling on it, which is also what would be reading on the balance.. -F F
Example: Gymnast on Rings What is the tension in each rope supporting a 150 lb gymnast if they make the angle of q relative to horizontal? Evaluate the tension at q = 30o. T1 T2 Fg
Example: Gymnast on Rings x: T1x + T2x = 0 y: T1y + T2y – Fg = 0 T1cosq – T2cosq = 0 T1 = T2T 2 T sin q = Fg = 150 lb T = 75 lb / sin q. At 30o, T = 150 lb. T1 T2 Fg
Pulleys • Pulleys can be used to redirect tension. Tension always follows the string. • If the pulley does not resist the motion of the string, it is said to be frictionless. • If the pulley is frictionless, and massless then the magnitude of the tension is the same on both sides. • Otherwise, T1 T2 . T1 T1 = T2 T2
Gymnast on Pulley • The gymnast weighs 150 lb. What is the tension on the rope? • T1 + T2 = Fg = 150 lb • T1 = T2 = 75 lb T1 T2 Fg
Atwood Machine • If the blocks have masses m1 = 6.0 kg and m2 = 4.0 kg, what is the acceleration of the blocks? • Assume the pulley is massless and frictionless, and neglect the mass of the string. m2 m1 What is the tension in the string?
Inclined Plane • A block slides down the inclined plane shown, starting at the top. How long does it take to reach the bottom? 5 m 3 m 4 m
Inclined Plane • Fnet= ma • Fnet = mg sin q = 3/5 mg • a = 3/5 g = 5.88 m/s2 • Time to reach bottom… • x = ½ a t2 • 5 m = (2.94 m/s2) t2 • t = 1.3 s. y N q Fnet x mg q sin q = 3/5 a = g/5 = 1.96 m/s2
Friction • If I push on an object and it doesn’t move, it means another force is pushing back to balance my force: staticfriction. • Fp = Ff Fp Ff
Friction • At some point, if I push hard enough the object will start to move. This limit to the amount of force friction can supply turns out to be proportional to the normal force. • Fp = Ff • Ff max = msN • ms N Fp coefficient of static friction Ff
Friction • Once the object is moving, the friction is called kinetic friction, and is again proportional to the normal force, but with a smaller coefficient. mk < ms • Ff max = mkN N Fp coefficient of kinetic friction Ff
Friction • Is the amount of friction affected whether I put the object on its long side or short side? • No • Is the amount of friction affected by how fast I push the object? • No Fp Fp
Friction • Suppose the coefficients of kinetic and static friction are mk = 0.7 and ms = 0.8, and the box weighs 700 N. It is at rest on a table. • If I push with a force of 400 N, what is the force of friction? • Ff = 400 N. • Fp < ms FN = 560 N Fp Ff
Friction • If I push with a force of 600 N, what is the force of friction? • Fp > msN = 560 N. • If I then reduce my force to 400 N, what is the force of friction? • Ff = 490 N until it stops. • Ff = mkN= 490 N Fp mg=700 N mk = 0.7 ms = 0.8 Ff
Inclined Plane with Friction If the board is tilted until the box just starts to slide, at an angle q= 37o. What is the coefficient of static friction? N Ff mg q
Inclined Plane with Friction It is convenient to pick axes so that two of the forces align with them. Decompose the weight into x and y components. The forces balance. x: Ff = mg sin q y: N = mg cosq ms= Ff/N = tan q = 0.75 x N Ff x mgcosq q q mg mg sin q q= 37o
Two Blocks • Two blocks of mass 5.0 kg each are being pulled at a constant speed by a 2.0 N force. What is the tension T in the rope connecting them? 5 kg T 2.0 N 5 kg
Two Blocks • What pulling force must be applied to accelerate the blocks at a rate of 0.2 g, and what is the tension in the rope? 5 kg T 2.0 N 5 kg