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Chapter02

# Chapter02 - Chapter 2 Forces and Motions 2.1 2.2 2.2 Forces...

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Chapter 2 Forces and Motions 2.1 Forces and Motions 2.2 A Block on a Wedge 2.2 Friction and Motions 2.1Forces and Motions (1) Newton’s first law Consider a body on which no net force acts. If the body is at rest, it will remain at rest. If the body is moving with a constant velocity, it will continue to do so. Inertial frame of reference : A non-accelerating frame of reference in which Newton’s first law is valid. (2) Newton’s second law The rate of change of momentum of a body is proportional to the resultant force and occurs in the direction of force. i.e. a m dt v d m F = . One Newton (1N) is defined as the force which gives a 1.0 kilogram mass an acceleration of 1.0 meter per second per second. Hence, if m = 1.0 kg and a = 1.0 ms -2 , then F = 1.0 N gives the proportional constant to equal one. dt v d m a m F = = F : vector sum of all forces F ma F ma F ma x x y y z z = = = Unit: International System of Units (SI units) F : Newton ( N ) m : Kilogram ( kg ) a : m/s 2 If F = 0 a = 0 or v = const. Newton’s first law 1

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B A 47 o m C mg 28 o knot (3) Newton’s third law : Whenever an object exerts a force on a second object, the second object exerts a force of equal magnitude but opposite direction on the first, i.e. BA AB F F - = force Reaction force Action For example, if a hammer exerts a force on a nail, the nail exerts an equal but oppositely directed force on the hammer. Note that the action and reaction forces come in pairs. They act on different bodies and so they will not cancel each other. Remark: A man stands on a platform scale. The reaction force from the scale to the man shows the weight of the man. Weight g m W = , Reaction g m R - = Note that mass m : a scalar measured in kg, weight W : a vector whose magnitude measured in N. Example 2-1 A block of mass m = 15 kg hanging from three ideal strings, find the tension in these cords. Answer: Equilibrium condition at knot: F = 0 x -direction: F Bx + F Ax = 0 or 0 28 cos 47 cos = - A B F F (1) y -direction: F Ay + F By + F Cy = 0 or 0 28 sin 47 sin = - + mg F F A B (2) From eqns.(1) and (2), we obtain F A = 104 N , F B = 135 N Remark: An ideal string has negligible mass and doesn’t stretch. An ideal pulley has negligible mass and is frictionless. It doesn’t change the magnitude of the tension from one side of the string to the other. 2
m 1 g cos 42 o m 1 g sin42 o N m 1 g T 42 o Example 2-2 Two blocks are connected by an ideal string passing over an ideal pulley as shown in the figure. The smooth inclined surface makes an angle of 42 o with the horizontal, and the block on the incline has a mass of m 1 = 6.7 kg. Find the mass of the hanging block m 2 that will cause the system to be in equilibrium. Answer: The free-body diagrams of the blocks m 1 and m 2 are shown as follows.

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