PS4 - Physics16 Spring2010 ProblemSet4 5 62 IDENTIF Y Apply...

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Physics 16 ʹ Spring 2010 ʹ Problem Set 4 5 . 62 . I DENTIF Y : Apply m F = a to each object . Constant speed means 0 a . S E T U P : The free-body diagrams are sketched in Figure 5.62. 1 T is the tension in the lower chain, 2 T is the tension in the upper chain and T F is the tension in the rope. E X E CUT E : The tension in the lower chain balances the weight and so is equal to w . The lower pulley must have no net force on it, so twice the tension in the rope must be equal to w and the tension in the rope, which equals F , is 2 w . Then, the downward force on the upper pulley due to the rope is also w , and so the upper chain exerts a force w on the upper pulley, and the tension in the upper chain is also w . E V A LUA T E : The pulley combination allows the worker to lift a weight w by applying a force of only /2 w . Figur e 5 . 62
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5 . 65 . I DENTIF Y : Apply m F = a to each block. S E T U P : Constant speed means 0 a . When the blocks are moving, the friction force is k f and when they are at rest, the friction force is s f . E X E CUT E : (a) The tension in the cord must be 2 m g in order that the hanging block move at constant speed. This tension must overcome friction and the component of the gravitational force along the incline, so 2 1 1 sin cos k m g m g and 2 1 (sin cos ) k m m . (b) In this case, the friction force acts in the same direction as the tension on the block of mass 1 m , so 2 1 k 1 ( sin cos ) m g m g , or 2 1 k (sin m m . ( c ) Similar to the analysis of parts (a) and (b), the largest 2 m could be is 1 s (sin cos ) m and the smallest 2 m could be is 1 s (sin cos ) m . E V A LUA T E : In parts (a) and (b) the friction force changes direction when the direction of the motion of 1 m changes. In part (c), for the largest 2 m the static friction force on 1 m is directed down the incline and for the smallest 2 m the static friction force on 1 m is directed up the incline. 5 . 66 . I DENTIF Y : A , to the hanging weight and to the knot where the cords meet. Target variables are the two forces.
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