Introduction to Newton's Laws of Motion

Transcription

1 Introduction to Newton's Laws of Motion

2 Outlook 1.The First Law on inertia states that every object will remain in a state of rest or uniform motion in a straight line unless acted upon by an external force. 2.The Second Law (F = ma) relates the cause (the force F) to the acceleration. Several different forces are discussed in the context of this law. 3.The Third Law states that action and reaction are equal and opposite.

3 Vectors r = i x(t) + j y(t) with ij-are the unit vectors in x and y Then, circular motion is then described as: r = i r sin(ωt) + j r cos(ωt) ω-is the angular velocity a=d 2 r/dt 2 = -ω 2 r or v 2 /r >this is the centripetal acceleration e r and e r is the unit radial vector a= -v 2 /r (e r ) Vector form ωt So any object that moves in a circle even at a constant velocity is accelerated!!

4 Inertia (dv/dt = a = 0) The First Law on inertia states that every object will remain in a state of rest or uniform motion in a straight line unless acted upon by an external force. -Everything remains at rest if no force is applied. -But everything will be at a uniform motion unless an external force acts on it? We know that for example things are going to stop after a time, a hockey disk travels from distance a to b and because of friction it will stop on Earth. What happens if there is an ideal smooth surface without friction? Will it travel for ever...can we check it in space? Earth is not perfectly inertial, because it has a small acceleration (see previous slide, it is accelerated slightly towards the Sun). But we will consider it for the moment a perfect inertial system i.e. not undergoing any acceleration. Examples of inertia, inertial frames and non inertial frames.

5 Second law: acceleration and force The Second Law (F = ma) relates the cause (the force F) to the acceleration. In other words, the acceleration of an object is directly proportional to the net force acting on. The net force determines the magnitude and direction of the motion of an object Force is a vector quantity F is a vector sum so that F = F1+F2+F Fn

6 Second law: acceleration and force F a ----force is proportional to the acceleration UNITS: F=ma force[newton], mass[kg], acceleration[m/s^2] in the (IS) (system, cgs) is a Dyne 1 dyne = 1 [(g.cm)/s^2] so what is 1N =?? Dyne Motion of planets as an example A good example is a string where F=-kx where k has units of N/m. Another is the gravitational force near the surface of Earth Fg=mg and friction F=μ.n The second law explains the motion of objects under the effect of a force!

7 Second law: acceleration and force acceleration and force add title Hooke's law F Δl F = -k Δl k=is the spring's constant or the restoring force

8 Third Law of Motion: Action and Reaction The Third Law states that action and reaction are equal and opposite. When two objects interact, the force exerted by the 1st object on the 2nd is equal in magnitude, but opposite in direction to the force exerted by the 2nd object on the 1st. F12=-F21 So what is happening in a system where F12=-F21=0? To take away: -Forces always occur in pairs -Each force in a force pair has the same magnitude and have opposite directions. -Each force in a force pair acts on a different system. The 3rd Law explains the source of the forces that cause the motion of objects.

9 Some fun

10 Some fun

11 Some fun What is here wrong?

12 Some Examples 1. PROBLEM Jose pulls a heavy box across the floor using a spring moves the box at a constant velocity, so that, spring is stretched 0.1 meters past its relaxed length. If the box has a mass of 35 kg and if the coefficient of friction between the box and the floor is mu=0.4, what is k-the spring constant? Hint: We know that Fr=mu.n and Fs=k.m (for friction and the spring) Solution: 1. diagram of forces m=35 kg μ=0.43 Fr x=0.1m k=? 35kg n Fs Jose mg

13 Some Examples 2. We need Newton's second law, and therefore we need to identify the forces acting in each component (this case x and y) For x 0 Σfx=ma_x = 0. Σfx=Fr-Fs=0 because they are opposite and Fr=μ.n and Fs=k.x k.x-μ.n =0 k=(μ.n)/x -> and n? For y 0 Σfy=ma_y = 0. Σfy=n-mg=0 So n=mg=(35).(9.8)= n=343 Newton Therefore substituting n in k=(μ.n)/x we get: k=( )/0.1 k=1372 N/m

14 Some Examples 2. PROBLEM Jose pushes a body that lies at an incline (15 deg). The initial velocity of the body is 2.m/s. We will assume there is no friction, so the body will go up and then slide down (like on ice or soap). The body has a mass of 25kg and we would like to know its acceleration a and how far it will go ( x). Solution: 1. diagram of forces m=25 kg Vo=2m/s x=? a=? 25kg n mg 15º

15 Some Examples 2. The second law applies again Then ΣF=ma So then Σfx=ma_x m.g sin(15º)=ma a=g sin(15º) a=2.54m/s 2 Now solve for x V 2 =Vo 2 +2a( x) Is the final sign positive or negative? mg cos(15) n mg sin(15)