Phys02

# Phys02 - Unit 2 Force and Motions 2.1 2.2 2.2 Force and...

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Unit 2 Force and Motions 2.1 Force and Motions 2.2 A Block on a Wedge 2.2 Friction and Motions 2.1 Force 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 v F m dt d m = . Now, one Newton is defined as the force which gives a mass of 1.0 kilogram 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 equals one. v Fa d mm dt == Σ F : vector sum of all forces Fm a a a xx yy zz = = = Unit: International system of units, or metric system SI F : Newton ( N ) m : Kilogram ( kg ) a : meter per second square (m/s 2 ) If Σ F = 0 a = 0 or v = const. Newton’s first law 1

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(3) Newton’s third law : Forces come in pairs, if a hammer A exerts a force on a nail B; the nail exerts an equal but oppositely directed force on the hammer. i.e. F A on B = F B on A Action force = Reaction force Note that they act on different bodies, so they will not cancel. Remark: A man stands on a platform scale. The reaction from the scale to the man shows the weight of the man. Weight W = m g , Reaction R = m g Mass m : a scalar, measured in kg Weight W : a vector, measured in N Example Two blocks are connected by a string 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. m 1 = 6.7 kg 42 o m 2 g T N m 1 g T m 1 g sin42 o N m 1 g T 42 o m 1 g cos 42 o m 2 The down-plane component of the 6.7-kg mass is given by N ms kg g m o o 9 . 43 42 sin ) 8 . 9 )( 7 . 6 ( 42 sin 2 1 = = 2
For equilibrium, this force is balanced if the tension has the same magnitude. On the other hand, the hanging block is also in equilibrium, the weight of it, m 2 g , must balance the tension force, hence we can write 9 . 43 2 = g m .

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## This note was uploaded on 09/06/2010 for the course BSC PHY1417 taught by Professor Prof during the Spring '08 term at HKU.

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Phys02 - Unit 2 Force and Motions 2.1 2.2 2.2 Force and...

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