Can you solve this problem on Mechanics?

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Answer 1 · 2016-11-13T15:23:17

Let the acceleration of the system be $a$ and tension on the connecting string be $T$

Considering the forces on $W_B$ we can write

$25xxgxxsin30-0.2xx25xxgxxcos30-T=25xxa.....(1)$

Considering the forces on $W_A$ we can write

$15xxgxxsin30-0.4xx15xxgxxcos30+T=15xxa.....(2)$

Adding (2) and (1) we get

$(25+15)xxgxxsin30-(0.2xx25+0.4xx15)xxgxxcos30=(25+15)xxa$

$=>(40xxgxxsin30-(5+6)xxgxxcos30=40xxa$

$=>40xxgxxsin30-11xxgxxcos30=40xxa$

$=>a=(40xx32xx1/2-11xx32xxsqrt3/2)/40$

$=16-4.4sqrt3=8.38fts^-2$

Taking $"acceleration due to gravity"g=32fts^-2$

Inserting the value of $a$ in (1)

$25xx32xxsin30-0.2xx25xx32xxcos30-T=25xx8.38$

$=>25xx32xx1/2-0.2xx25xx32xxsqrt3/2-T=25xx8 38$

$=>400-80xxsqrt3-T=209.5$

$=>261.44-T=209.5$

$=>T=261.44-209.5=51.94poundal$