Question

Using the coefficients of static friction shown, determine the smallest mass of block A so that it will prevent sliding motion of B down the plane.

A cable is attached to the 20-kg plate B, passes over a fixed peg at C, and is attached to the block at A. (Figure 1) Figure < 1 of 1 > Mc = 0.3 MA = 0.2 MB = 0.3 30°

Answer 1

Drawing FBD of the system, From FB, For Block A, ing T Efxzo fa stand ti-baqa) - mg como = 0_0 y (mg) on splitting . Ti-CMANA)- ma sin na the {fyzo NA -mg cose 20. -- . For plate B, . EFX20 Tz- mag sino + MB No + MANA 20. — Efy 20 . Ng - NA - Meg covo. 20 – .:. ③ For Pegc, Te Men Mc Tr. . Mc 20.3 0230? giveri MB2 20 kg ; MA202, MB 20,3 on substituting, equalion @) NA - MA *9.81.CO 3020 - TNA: 8.49 MA] en (1) - I - 0.2 x 8.49 MA - MAX 9.81 sin 30° - - T1 = 6.603 MA | eqi (s) a. Izz tie Mell 26.603 MA eos 0.37 . 2(6.603 x 2.566) MA... T Tzz 16.9 MAI

ernation (3) - 16.9MA - 2089.81 sig30 +0.3 No to. 0.248.49 MA 20. 1 0.3 Ng = 98.1 -18.598 MA.-@ Foom equ (4) No - 8.49 IMA – 2049.81 x colo' 20. on Multiply 0.3 : 0.3 Ng 2 2.547 MA + 50.973 —O. ewalion @ - equation @. a 21.145 MA 2 47.127. I MA 2 2.22 kg. I

the smallest mass of block A so that it will prevent sliding motion of B down the plane is 2.22 Kg.

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