A) 20 N
B) 40 N
C) 10 N
D) 32 N
Correct Answer: D
Solution :
[d] For equilibrium of all 3 masses, \[{{T}_{3}}=({{m}_{1}}+{{m}_{2}}+{{m}_{3}})\]a or \[a=\frac{{{T}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] For equilibrium of \[{{m}_{1}}\] & \[{{m}_{2}}\] \[{{T}_{2}}=({{m}_{1}}+{{m}_{2}}).a\] or, \[{{T}_{2}}=\frac{({{m}_{1}}+{{m}_{2}}){{T}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] Given \[{{m}_{1}}=10kg,\]\[{{m}_{2}}=6kg,\]\[{{m}_{3}}=4kg,\] \[{{T}_{3}}=40\,N\] \[\therefore \]\[{{T}_{2}}=\frac{(10+6).40}{10+6+4}=32N\]You need to login to perform this action.
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