T1a4s - Assignment 4 solutions (Fall term Physics 102) 4.37...

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4.37 (a) Since the crate has constant velocity, 0 xy aa = = . Applying Newton’s second law: , or cos 20.0 0 xk x Assignment 4 solutions (Fall term Physics 102) FF fm a Σ= ° − = = ( ) 300 N cos20.0 282 N k f = °= and , or sin 20.0 0 y FnF w Σ=− ° −= ( ) 3 300 N sin 20.0 1000 N 1.10 10 N n + = × . The coefficient of friction is then 3 282 N 1.10 k k f n µ = == × 0.256 . (b) In this case, , so . The friction force now becomes sin 20.0 0 y w Σ=+ ° sin 20.0 897 N nwF =− ° = ( )( ) 0.256 897 N 230 N kk fn = . Therefore, cos 20.0 x kx w x F Ff m a g ⎛⎞ ° = ⎜⎟ ⎝⎠ a and the acceleration is ( ) ( ) [ ] ( ) 2 300 N cos 20.0 230 N 9.80 m s cos 20.0 1000 N k g a w °− = 2 0.509 m s 4.41 The normal force acting on the crate is given by cos nFm g θ =+ . The net force tending to move the crate down the incline is sin s mg f , where s f is the force of static friction between the crate and the incline. If the crate is in equilibrium, then sin 0 s f , so that sin s g = . But, we also know ( ) cos ss s fn F m g µθ ≤= + . mg F n fs θ = 35.0° Therefore, we may write ( ) sin cos s F ≤+ , or () ( ) 2 sin sin 35.0 cos cos 35.0 3.00 kg 9.80 m s 0.300 s Fm g ° ≥− = − ° = 32.1 N 4.58
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This note was uploaded on 01/15/2011 for the course PHYS 102 taught by Professor Vandenberg during the Fall '08 term at University of Victoria.

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T1a4s - Assignment 4 solutions (Fall term Physics 102) 4.37...

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