kinetic - 43 Chapter 2. Kinetics 2.1 Rectilinear Motion...

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43 Chapter 2. Kinetics 2.1 Rectilinear Motion (Newton’s law) Newton’s second law states that the total force on a particle is equal to the rate of , if is constant or dd V F mV m m F ma =→ = G GG G G change of its linear momentum, which is the product of its mass and velocity: ( ) , dt dt A reference frame in which Newton’s second law is valid is called Newtonian or inertia reference frame. The earth is, strictly speaking, not Newtonian, but approximately so for the purpose of general engineering excluding aerospace applications. A reference frame moving at constant velocity relative to earth surface is Newtonian, while an accelerating or rotating one is not. For example, a person standing on a scale in an upward accelerating lift experiences heavier eight but he does not have relative lative to a weight, but he does not have relative a relative to the lift. Newton’s second law fails inside this lift. Ground fixed reference frame: N-W=m a Lift fixed reference frame: N-W =0 (incorrect result – non-inertial frame)
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44 2.1 Rectilinear Motion (mass centre) For a collection of 1 particles, each particle satisfies the second law , where is the force exerted by particle on particle . ii j i j iN F fm a f j i =→ += GG G G and is the external force acting on part jj ji i F G icle . i Newton's 3rd law states that . Therefore, the sum of the second law expressions for all particles leads to cancellation of all interaction forces: ff =− G G 2 2 , where . De i i i dr Fm a a dt == ∑∑ G G 2 fine the vector of mass centre as G ( ) 2 , where , we have ,, , 1 i i x yy zz x y z c ci i i c d r rm r F m a mm FF e F e F e F F F mt m d = ∑= + + = = G GGG G G () , , xx xyz aa a a F m a =∑ = G xy z z a a a
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45 2.1 Rectilinear Motion (ex. 1) Pre-requisite knowledge: friction is calculated as friction coefficient times the normal support force by the ground, which is normally not equal to the object weight mg nless it’s placed horizontally and is not accelerating in the vertical direction Question : The 2 crates are released from rest. Their masses are m A = 0 kg nd = 30 kg and the coefficients of friction between crate A unless it s placed horizontally and is not accelerating in the vertical direction. 40 kg and m B 30 kg , and the coefficients of friction between crate A and the inclined surface are μ s = 0.2 ( static ) and μ k = 0.15 ( dynamic ). What is the acceleration of the crates? Strategy We must first determine whether A slips . We ssume that the crates remain stationary assume that the crates remain stationary and see whether the force of friction necessary for equilibrium exceeds the aximum friction force maximum friction force. If slip occurs, we can determine the resulting acceleration by drawing free-body diagrams of the crates and applying Newton’s 2 nd law to them individually.
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46 2.1 Rectilinear Motion (soln 1) Solution : if it does not slip, tension T=weight of crate B.
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kinetic - 43 Chapter 2. Kinetics 2.1 Rectilinear Motion...

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