Newton S Laws Of Motion

DYNAMICS
-studies the relationship of motion to the forces that causes it.
Types of Forces:
(a) Normal Force, n :When an object rests or pushes on a surface, the surface exerts a push on it that is directed perpendicular to the surface.
(b) Friction Force, f : In addition to the normal force, a surface may exert a frictional force on a object, directed parallel to the surface and opposite the motion or impending motion of the object. f s = µ s n - static friction, maximum friction before the object begins to move.

n ff kk ==µµk k n - kinetic friction, friction on a moving object.

DYNAMICS

Types of Forces:
(c) Tension Force, T : A pulling force applied on an object by any longitudinal object. Along the longitudinal object, tension is always directed away any point of consideration.
(d) Compression Force, C : A pushing force applied on an object by any longitudinal object. Along the longitudinal object, compression is always directed towards any point of consideration.
(e) Weight, W : The pull of gravity on an object. It is always directed vertically downward. W = mg

point of consideration point of consideration NEWTON’s LAWS OF MOTION

F =0

F
F=
0
=0

F =0

F
FFxxx ==
0
=00

=000
FFyx ==
F
Fyy = 0
Fy = 0

First Condition of
Equilibrium

NEWTON’s LAWS OF MOTION

Example 1. A large wrecking ball is held in place by two light steel cables as shown in the figure. If the mass m of the wrecking ball is
4,900 kg, what are (a) the tension TB in the cable that makes an angle of 40o with the vertical and (b) the tension TA in the horizontal cable?

NEWTON’s LAWS OF MOTION

Example 2. A farmer hitches her tractor to a sled loaded with a firewood and pulls it along a level ground with constant speed. The total weight of the sled and the load is 14,700N. The tractor exerts a force on the sled at an angle of 36.9o above the horizontal. If the coefficient of kinetic friction between the sled and the ground is
0.333, find (a) the force exerted by the tractor on the sled; (b) the friction and normal force on the sled.

NEWTON’s LAWS OF MOTION

NEWTON’s LAWS OF MOTION

F =0
F
F=
=00ma
ma ma Fx =ma
0
F
0
=
F
0 ma =
Fxx = 0ma xxx ma y ma 0
=
F
F
0 ma =
Fyx = 0 yy
Fyy = 0
Fy = 0

NEWTON’s LAWS OF MOTION

Example 3. In a truck-loading station at a post office, a small 0.200-kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m, as shown below. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.
Determine (a) the acceleration of the package; (b) the kinetic coefficient of friction between the package and the horizontal surface;
(c) friction and the normal force.

NEWTON’s LAWS OF MOTION

Example 4. Consider the system in the figure; the pulley is frictionless.
Block A weighs 45.0 N and block B weighs 25.0 N. Once block B is set into downward motion, calculate (a) the tension in the string if the kinetic coefficient of friction between block A and the horizontal surface is 0.20; and (b) acceleration of the two blocks.

NEWTON’s LAWS OF MOTION

FB on A

FA on B

FA on B = -FB on A

NEWTON’s LAWS OF MOTION

Example 5. A 100-N crate is hanging on a cable supported by strut with negligible mass. Find the tension T in each cable and the magnitude and direction of the force exerted on the strut by the pivot as shown in the figure.

100 N