Since the weight of block Ais 44 N, the least weight for Cis (110 – 44) N = 66 N. (b) The second law equations become T– f= (WA/g)aFN– WA= 0 WB– T= (WB/g)a.In addition, f= μkFN. The second equation gives FN= WA, so f= kWA. The third gives T=WB– (WB/g)a. Substituting these two expressions into the first equation, we obtain WB– (WB/g)a– kWA= (WA/g)a.Therefore,()()22(9.8 m/s ) 22 N0.15 44 N2.3 m/s .44 N + 22 NBkAABgWWaWW−−===+31. (a) Free-body diagrams for the blocks Aand C, considered as a single object, and for the block Bare shown below. Tis the magnitude of the tension force of the rope, FNis the magnitude of the normal force of the table on block A,fis the magnitude of the force of friction, WACis the combined weight of blocks Aand C(the magnitude of force GFgACshown in the figure), and W
This is the end of the preview. Sign up
access the rest of the document.
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.