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Physics 104

Outline - Newton's Laws

  1. Newton's Laws

    1. First Law. An object at rest remains at rest and an object moving with velocity v remains moving with velocity v if and only if no net external force acts on the object.

    2. Second Law. The net force acting on an object is directly proportional to and in the same direction as the acceleration. The proportionality constant is the mass of the object. Fnet = ma

    3. Third Law. When two objects interact in whatever manner, the force F12 exerted by the first object on the second is equal in magnitude and opposite in direction to the force F21 of the second object on the first. - F12 = F21

  2. Application of Newton's Laws

    1. An object hanging motionlessly in Fig. 1 above has a weight of 800 N. Find the tensions T1 and T2 in the massless ropes. For the object to be at rest, the sum of the forces acting on the object must be zero. Since the tensions are neither in the X or Y-direction, take components:

      (Fnet)x = max = 0

      (Fnet)y = may = 0

      T2 cos 20o – T1 cos 20o = 0

      T2 sin20o + T1 sin20o – 800N = 0

      T2 cos 20o = T1 cos 20o

      T2 sin20o + T1 sin20o = 800N

      T2 = T1 = T     2T sin20o = 800 N     T = 1.17 x 103 N = T2 = T1

    2. When you apply Newton’s second law, you must be very careful about your choice of system.

      1. Imagine two masses m1 = 1.0 kg and m2 = 2.0 kg are connected by a massless string and are pulled along a frictionless surface by a force F = 9.0 N as shown in Fig. 2a below. Find (a) the acceleration of the blocks and (b) the tension in the string.

        1. In Fig. 2a above, I have isolated the entire system of the two blocks. There are vertical forces acting on both blocks: the normal force of the surface and the weight of each block. Since there is no acceleration in the vertical direction and there is no friction (we would need to know the normal force in order to find the frictional force on a block), we shall not consider forces in the vertical direction. The only external horizontal force acting on the system is F = 9.0 N. There are internal horizontal forces acting on the blocks and the string, but we are not interested in internal forces because
          Fnet external = msystem a
          9.0 N = [(1.0 + 2.0)kg] a,
          and   a = 3.0 m/s2

        2. Now we choose our system as shown in Fig. 2b and 2c above.  Again I only look at the horizontal forces. I have isolated:

          1. the string experiencing F1s, the force of block 1 on the string and F2s, the force of block 2 on the string

          2. m1 experiencing Fs1 = the force of the string on block 1

          3. m2 experiencing Fs2, the force of the string on block 2 and F = 9.0 N.

        3. For the string, Fnet external = mstringa
          F2s - F1s = (0)3.0 m/s2 = 0
          So F2s = F1s
          By Newton's Third Law, Fs2 = F2s and Fs1 = F1s
          Thus, Fs2 = F2s = F1s = Fs1.
          We drop all this fancy notation and write
          Fs2 = F2s = F1s= Fs1 = T as shown in Fig. 2c above.

        4. For m1,  Fnet external  = m1a

          T = 1.0 kg(3.0 m/s2) = 3.0 N

          For m2,  Fnet external  = m2a

          9.0 N - T = 2.0 kg(3.0 m/s2) = 6.0 N

          9.0 N - 6.0 N = T = 3.0 N

    3. As you can see from the above example, Newton Third Law of Motion forces always act on different objects. In other words, they do NOT act on the same object.

    4. Practice problems in 104 Problem Set for Newton's Laws: 1 - 8.

  3. Frictional Forces

    1. The frictional force f = µN, where the coefficient of friction µ is labeled with a subscript s for static situations and k for objects in motion. The coefficient of friction for static situations is always greater than that for kinetic. The static frictional force can go from 0 to a value necessary to just about move the object.

    2. Finding the frictional force (Fnet)y = may = m(0) = 0

      1. For Fig. 3a above, N – mg = 0. N = mg and f = µN = µmg
      2. For Fig. 3b, N + F sin Θ – mg = 0. N = mg - F sin Θ and f = µN
        = µ(mg - F sin Θ)
      3. For Fig. 3c, N - mg cos Θ= 0. N = mg cos Θ; f = µN = µmg cos Θ

    3. Practice problem in 104 Problem Set for Newton's Laws: 10.

  4. Free Body Diagrams

    1. Force Diagrams such as those shown in Fig. 4 below are called Free-body diagrams. They are extremely important in solving Newton Second Law of Motion problems.

    2. In each situation shown in Fig. 4a above, one or more forces act upon an object. All drawings are in a vertical plane and friction is negligible except in (b) and (d). Draw free body diagrams for the figures, scale the forces as close as possible. Label all the forces acting on the objects. If the object has an acceleration, show its direction. If there is no acceleration, indicate that it is zero.

    3. In Fig. 4b above, (a) the only force that acts on the object is its weight mg. Its Acceleration a is down. (b) For a constant velocity, the net force acting on the object must be zero. The upward frictional force equals the weight of the object. (c) There is no net force perpendicular to the plane. The normal force is equal to the component of the weight perpendicular to the plane. The component of the weight parallel to the plane gives the acceleration down the incline. (d) Now in addition to the forces talked about in (c) there is a frictional force up the plane that is equal in magnitude to the component of the weight parallel and down the plane. There is no acceleration. (e) and (f) The only force acting on an object in projectile motion (neglecting a frictional force) is the weight of the object down.

    4. Practice problems in 104 Problem Set for Newton's Laws: 10 - 14, 16, 18, and 21.

  5. Uniform Circular Motion

    1. An object has uniform circular motion when it moves in a circle with constant speed. The speed v is constant, but the velocity v is not constant because v is always tangent to the path so it continually changes direction. In Fig. 5 below, |v1|= |v2| = |v3| = |v4| = v, but this is not a case of constant velocity because the direction of v changes.

    2. In Fig. 6 below, the two angles labeled Θare equal because v2 an v1 are perpendicular to their respective radii. Since |v1 | = |v2| = v, the triangle with the radii and the triangle with the velocities are similar because they are isosceles triangles and they have angles that are equal. Thus Δv/v = Δr/r or Δv = Δr(v/r). Divide both sides by Δt and take limit as Δt approaches zero

      (a) = (v)(v/r) or a = v2/r. Since a = dv/dt, a has the same direction as Δv or into the center of the circle.

      1. The magnitude of the acceleration is constant: |a1| =
        |a2| = |a3| = |a4| = a = v2/r, but this is not a case of constant acceleration because the direction of a is not constant.

      2. Since the acceleration is into the center, for a centripetal acceleration there must be a net force into the center. This force is perpendicular the to the tangent or the direction of the velocity. There is no component of the force in the direction of the velocity and, therefore, there is no change in the magnitude of the velocity.

      3. Descriptive Terms

        1. Period T = time for one complete rotation = 2 πr/v
        2. Frequency f = reciprocal of period = 1/T = v/2 πr

    3. Practice problems in 104 Problem Set for Newton's Laws: 15, 17.

  6. Gravitational force Fg = Gm1m2/r2, where m1 and m2 are the masses of two point particles separated by a distance r.

    1. Near the earth’s surface at height h very small compared to the radius of the earth RE.

      1. You can treat a spherical mass ME as a point mass located at the center of the earth. The force on an object of mass m at height h above the earth by the earth is Fg = GmME/(RE + h)2. Since RE is much greater than h, drop the h and Fg = GmME/RE2.In general, Fnet = ma

      2. For a freely falling object, GmME/RE2 = ma
        For free fall a = g         GmME/RE2 = mg = Weight of object

      3. (mg) is not a mass times an acceleration, it is a force. To determine the direction of the force in a problem, decide on a direction to call positive and then see if mg is in that direction. After doing this, never substitute –9.8 m/s2 for g in (mg).

      4. Practice Problem I. A classics student of mass 70.0 kg is 0.50 m from a physics student of mass 50.0 kg. Calculate the approximate magnitude of the gravitational force that each exerts on the other. Explain why your result is only approximately equal to the magnitude of the actual gravitational force.

        Solution. The approximate magnitude of the gravitational force that each exerts on the other is found from the equation for point-like masses.

        F = Gmcmp/r2
           = (6.67 x 10-11 N-m2/kg2)(70.0 kg)(50.0 kg)/(0.50 m)2
           = 9.3 x 10-7 N.

        The force is approximate because students are not point-masses.

      5. Practice Problem II. An astronaut puts a bowling ball into a circular orbit about the Earth at an altitude h of 350 km. Find the ball's period of motion. The radius and mass of the earth are M = 5.98 x 1024 kg and R = 6.37 x 106 m, respectively.

        Solution. The radius of the orbit of the bowling ball r = the radius R of the earth + the height h above the earth's surface =
        (6.37 + 0.35)106 m,
        and the constant in Newton's law of gravitation G =
        6.67 x 10-11 N-m2/kg2.
        The gravitational force produces the centripetal acceleration:

        GMm/r2 = mv2/r(1)

        GMm/r2 = m(2πr/T)2/r

        Since v = 2 πr/T,  or T2 = 4π2 r3/GM,  and

        T = 2 π (r3/GM)1/2

        = 2{(6.72 x 106 m)3/6.67 x 10-11 N-m2/kg2 x 5.98 x 1024 kg}1/2
        = 55 x 102 s


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Susan D. Kunk
Phyllis J. Fleming
September 25, 2002
April 2, 2003