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Unformatted text preview: 528Chapter 5(b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that each hook exerts, which is the same as the horizontal component of the force due to the tension at the end; endmiddlecos,TT=so middlecos(2sin )(2tan ).TMgMg==(c) Mathematically speaking, because this would cause a division by zero in the equation for endTor middleT. Physically speaking, we would need an infinite tension to keep a nonmassless rope perfectly straight. EVALUATE:The tension in the rope is not the same at all points along the rope. 5.64.IDENTIFY:Apply m=Fa!!to the combined rope plus block to find a. Then apply m=Fa!!to a section of the rope of length x. First note the limiting values of the tension. The system is sketched in Figure 5.64a. At the top of the rope TF=At the bottom of the rope ()TM ga=+Figure 5.64a SET UP:Consider the rope and block as one combined object, in order to calculate the acceleration: The freebody diagram is sketched in Figure 5.64b. EXECUTE:yyFma=()()FMm gMm a+=+FagMm=+Figure 5.64b SET UP:Now consider the forces on a section of the rope that extends a distance xL&lt;below the top. The tension at the bottom of this section is ( )T xand the mass of this section is ( / ).m x LThe freebody diagram is sketched in Figure 5.64c. EXECUTE:yyFma=( )( / )( / )FT xm x L gm x L a=( )( / )( / )T xFm x L gm x L a=Figure 5.64c Using our expression for aand simplifying gives ( )1()mxT xFLMm=+EVALUATE:Important to check this result for the limiting cases: 0:x=The expression gives the correct value of .TF=:xL=The expression gives (/()).TF MMm=+This should equal (),TM ga=+and when we use the expression for awe see that it does. 5.65. IDENTIFY:Apply mF = a!!to each block. SET UP:Constant speed means a=. When the blocks are moving, the friction force iskfand when they are at rest, the friction force is sf. EXECUTE:(a)The tension in the cord must be 2m gin 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 ( )211sincoskm gm gm g=+and 21(sincos )kmm=+. (b) In this case, the friction force acts in the same direction as the tension on the block of mass 1m, so 21k1(sincos)m gm gm g=, or 21k(sincos )mm=. Applying Newton&amp;s Laws 529 (c)Similar to the analysis of parts (a) and (b), the largest 2mcould be is 1s(sincos )m+and the smallest 2mcould be is 1s(sincos )m. EVALUATE:In parts (a) and (b) the friction force changes direction when the direction of the motion of 1mchanges. In part (c), for the largest 2mthe static friction force on 1mis directed down the incline and for the smallest 2mthe static friction force on 1mis directed up the incline....
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Force

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