A block of mass 20kg is suspended through two light spring balances as shown in figure. Calculate the reading of spring balance (1) and (2) respectively.
A
200N,400N
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B
400N,200N
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C
0N,200N
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D
200N,200N
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Solution
The correct option is D200N,200N FBD of each component of spring+block system:
From FBD of block, applying equilibrium condition in vertical direction, T2−mg=0 ∴T2=mg=20×10=200N...(i)
Since spring balances are assumed massless and system is in equilibrium i.e a=0 For spring balance 2, T2−T1=m2×a=0×0=0 ∴T1=T2=200N Simillarly, for spring balance 1, T′−T1=m1a=0×0=0 ∴T′=T1=200N
Therefore, reading of both spring balances will show 200N