Newton’s Laws of Motion. We begin our discussion of dynamics (study of how forces produce motion in objects) We’ll use kinematic quantities (displacement, velocity, acceleration) along with force & mass to analyze principles of dynamics
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Draw all the external forces acting on the hot dog cart
Weight of cartNewton’s Laws of Motion
(in this case)
v = constant a = 0Newton’s Laws of Motion
(Newton’s 1st Law)
CQ2: If F is the force of air resistance on an object with mass m moving at a constant velocity, which of the following best describes the acceleration of the object when the force of air resistance is reduced by a factor of 4?
Which animation correctly shows the motion of the satellite after the thruster force is applied?
PHYSLET #8.2.2, Prentice Hall (2001)
A setup similar to the one shown at right is often used in hospitals to support and apply a traction force to an injured leg. (a) Determine the force of tension in the rope supporting the leg. (b) What is the traction force exerted on the leg? Assume the traction force is horizontal.
m2gExample Problem #4.36
Find the acceleration of each block and the tension in the cable for the followingfrictionless system:
Coupled system: mass m2 moves same distance in same time as mass m1 v1 = v2 a1 = a2 = a
Apply Newton’s 2nd Law to block m1:
S Fx = m1ax T = m1ax= m1a (1)
SFy = 0 m1g – N = 0 m1g = N (a = 0 in y–direction)
Apply Newton’s 2nd Law to block m2:
S Fy= m2ay = m2a
m2g – T = m2a (2)
Combining equations (1) and (2):
m2g – m1a = m2a a = [m2 / (m1 + m2)]g = 6.53 m/s2
Using equation (1) and plugging in for a:
T = m1a = 32.7 N
(Note that this analysis will be useful for Experiment 5 in lab!)
What is the maximum height reached by the rocket?
ActivPhysics Problem #2.4, Pearson/Addison Wesley (1995–2007)
is always to
(pushing but no sliding)
(Note that f is in the same direction as the motion of the car, but opposite to the motion of the tires!)
A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck’s flatbed is 0.350, and the coefficient of kinetic friction is 0.320. (a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck’s flatbed? (b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?
f1Example Problem #4.69
Two boxes of fruit on a frictionless horizontal surface are connected by a light string (see figure below), where m1 = 10 kg and m2 = 20 kg. A force of 50 N is applied to the 20-kg box. (a) Determine the acceleration of each box and the tension in the string. (b) Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10.
Solution (details given in class):
(a) a = 1.7 m/s2, T = 17 N
(b) a = 0.69 m/s2, T = 17 N
( f1and f2 = 0 in part a)