Chapter 5 Further Applications of Newtons Laws
5-1 Friction When we try to slide an object across another surface, the microscopic bumps impede the motion. When a body is in motion along rough a surface, the force of friction acts opposite to the direction of the bodys velocity. The magnitude of the force of kinetic friction depends on the nature of the two sliding surfaces. For a given surface, experiment shows that the force is approximately proportional to the normal force between the two surfaces.
Static Friction vs. Kinetic Friction
Conceptual Example 5-2
To push or pull a sled? Your little sister wants a ride on her sled. If you are flat on the ground, will you exert less force if you push her or pull her? Assume the same angle in both cases.
Example 5-4
Pulling against friction. A 10.0-kg box is pulled along a horizontal surface by a force FP of 40.0 N that is applied at a 30.0o angle. We assume a coefficient of kinetic friction of 0.30. Calculate the acceleration.
Example 5-8
A plane, a pulley, and two boxes. A box of mass m1 = 10.0 kg rests on a surface inclined at = 37o to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box of mass m2, which hangs freely. (a) If the coefficient of static friction s = 0.40, determine what range of values for m2 will keep the system at rest. (b) If the coefficient of kinetic friction is k = 0.30, and m2 = 10.0 kg, determine the acceleration of the system.
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The rope is massless and taut, so the tension in the rope is equal to F. 1. F F f 1 = ma 2. F + F f 1 F f 2 = Ma 3. N1 = mg 4. N 2 = mg + Mg 5. F f 1 = sN1 = smg 6. F f 2 = k N 2 = k (m + M)g
from 1 and 5 7. F - smg = ma from 2, 5 and 6 8. F + smg k (m + M)g = Ma F - smg + F + smg k (m + M)g = ma + Ma (add 7 and 8) 2F k (m + M)g = (m + M)a 2F (m + M)a = k (m + M)g 2F 1 k = ( a) m+M g F a 2F 1 F 2F 1 F M m s k = ( a) = = mg g m + M g mg m + M g mg M + m
F - ma = smg (7) F a s = mg g
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5-2 Uniform Circular Motion
An object that moves in a circle at constant speed v is said to undergo uniform circular motion. v2 aR =
The velocity vector and the acceleration vector are always perpendicular to each other.
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Frequency and Period
Circular motion is often described in terms of the frequency f as so many revolutions per second. The period T of an object revolving in a circle is the time required for one complete revolution. Period and frequency are related by:
T=
f
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5-3 Dynamics of Circular Motion
v2 FR = maR = m r
[circular motion]
Centripetal force
Centrifugal force
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Problem Solving Uniform Circular Motion
1. Draw a free-body diagram, showing all the forces acting on each object. Be sure you identify the source of each force. 2. Determine which of these forces act to provide the centripetal acceleration. The sum of the radial forces (or components) provides the centripetal acceleration, aR = v2/r.
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Conical pendulum. A small ball of mass m, suspended by a cord of length L, revolves in a circle of radius r = L sin, where is the angle the string makes with the vertical. (a) In what direction is the acceleration of the ball and what causes the acceleration? (b) Calculate the speed and period, T, of the ball in terms of L, , and m.
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5-4 Highway Curves, Banked and Unbanked
The car follows a curved path. By Newtons 1st law, your body wants to go straight. A force is exerted on you to make you also follow a curved path. The force is exerted by the friction of the seat or by direct contact with the door.
Static friction.
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Example
Skidding on a curve. A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 50 km/h (14 m/s). Will the car make the turn, or will it skid, if: (a) the pavement is dry and the coefficient of static friction is s = 0.60; (b) the pavement is icy and s = 0.25?
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Example
At what minimum speed must a roller coaster be traveling when upside down at the top of a circle if the passengers are not to fall out? Assume a radius of curvature of 8.0 m.
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*5-5 Nonuniform Circular Motion
Speed of an object is changing if the force has a tangential component.
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A block of mass m slides along a horizontal surface lubricated with a thick oil which provides a drag force proportional to the square root of velocity: FD=-bv1/2 a. If v = v0 at t = 0, determine as function of time. b. Determine x as function of time. Assume that x = 0 at t = 0. c. Determine the maximum distance the block can travel.
v0 FD m x=0 t=0 m
vf = 0 x = xf t = tf
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