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Calculate the net acceleration produced in the arrangement shown below. |
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Answer» Solution :(a) Net pulling force =3 mg-mg=2MG Total mass =3m+m=4M `a=(2mg)/(4m)=(g)/(2) ` (B) Pulling force= 2mg Total mass =3m `a=(2 mg)/(3m)=(2g)/(3)` (c ) Net pulling force= 3mg-2mg=mg Total mass =3m+m+2m=6 m `a=(mg)/(6mg)=(g)/(6)` (d) Net pulling force= 3 mg +mg-2 mg= 2 mg Total mass =3m+m+2m=6m `a=(2g)/(6mg)=(g)/(5)` (e) Net pulling force = 4 mg `sin 30^(@)-mg`=mg Total mass =4m+m= 5m `implies a=(mg)/(5m)=(g)/(5)` |
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