Force and Free Body Diagram Practice Questions - The Burns Home ...

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6.T 3T 1T 351T 2T 13551T 2T 235W 2 W 1W 2400N(a) (b) (c)From (c) T 1sin 35400NFrom (b) T cos513 2400NT1sin 35697.39NTT2T3cos 51571.26Ncos 51907.74NFrom (c) T2 T1cos35 N 697.39 cos 35571.26NFrom (b) T sin 51907.74Nsin 51W3 2W22W 705.45NT 697 N, T 571 N, T 908 N, W 705NTherefore :1 2 3 27.AB30F fF NF TF TF NF gF fmg cos30mgsin 30 30mgF gFf A FN A0.15mg m 0.1540kg9.82s 58.8NFf B FN B0.15 mg cos 30 m 0.1540kg9.8 0.8660252s50.922NWe need only applyx xFmafor each diagram, since masses are moving (unbalanced)
F F maTAf A xF 58.8N 40kg aTAF 40kg a 58.8NTAxxmg sin 30F F maTBf B x196N F 50.922N 40kg aTBF 40kg a 145.078NTBxxTherefore x x80kg a86.278N40kg a 58.8N 40kg a 145.078NaAlso xx89.722N80kgm1.0782sT 40kg a 58.8Nx m 40kg1.078 58.8N2 s 101.939NTherefore the masses have an acceleration of108 m s 2and a tension of 102N8.10 kg6 kg9 kgLetLetTLbe the tension in the rope on the leftTRbe the tension in the rope on the rightNow since the system is moving we will choose downward as positive in 9kg block, right as positivein the 10kg block, and up as positive in the 6kg block.For the 9kg block we have :For the 6kg block we have :For the 10kg block we have :Adding all three equations F T m ag9 R 9T F m aL g 6 6TR T LF f 10m 10ag9 g6 f10 6 9 1088.2N 58.8N 0.2 10kg m 9.8 25kg a2 s ma 0.3922sF F F m m m aThe acceleration is 0.39 m 2sNow, Subbing acceleration into the first equation to findT R
Page 1 and 2: Forces Questions1. Draw a Free Body
Page 3 and 4: Solutions1.Question 1y50T 240T2 sin
Page 5: TTBy symmetry 4 34T T4 1 N s

Force and Free Body Diagram Practice Questions - The Burns Home ...

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