Tmmgmm=+FHGIKJ−=2112105( N.μμθ)cos.(b) These equations lead to an acceleration equal to ag=−++FHGIKJFHGIKJ=sincos..2211362 m/s2(c) Reversing the blocks is equivalent to switching the labels. We see from our algebraic result in part (a) that this gives a negative value for T(equal in magnitude to the result we got before). Thus, the situation is as it was before except that the rod is now in a state of compression. 68. The free-body diagrams for the two boxes are shown below. Tis the magnitude of the force in the rod (when T> 0 the rod is said to be in tension and when T< 0 the rod is under compression), 2NFGis the normal force on box 2 (the uncle box), 1NFGis the the normal force on the aunt box (box 1), Gf1is kinetic friction force on the aunt box, and Gf2is kinetic friction force on the uncle box. Also,
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.