A bob of mass ′m′ is attached to ceiling of a lift with the help of string as shown. Calculate tension in string in equilibrium position, when lift is accelerated forward with acceleration ′a′.
A
m(g+a)
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B
m(g−a)
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C
m√g2+a2
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D
mg
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Solution
The correct option is Cm√g2+a2 As lift has forward acceleration pseudo force will be backwards hence FBD is
⇒ here as mg and ma are perpendicular to each other. Net force of mg,ma is balanced by T So, T=√(ma)2+(mg)2 T=m√g2+a2