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Three masses m1,m2,m3 are connected with string and placed on smooth table. Now m1 is pulled with force F and common acceleration produced is a. Find individual force on each mass.

Three masses m1,m2,m3 are connected with string and placed on smooth table. Now m1 is pulled with force F and common acceleration  produced is a. Find individual force on each mass. 

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Grade:12th pass

1 Answers

Susmita
425 Points
5 years ago
a=F/(m1+m2+m3)
For each box weight is balanced by tension.
For m3,
T3=m3a=m3F/(m1+m2+m3). This is the force on m3.
Due to equality between action and reaction T1 will act on m2 in left direction.Also T2 will act in the right.So
T2-T1=m2a
or,T2=m2a+T1.Put the values obtained earlier ti get T2.This is the force on m2.
In case of m1 also T2 acts in left and T3 acts in right.So
T3-T2=m1a
T3=T2+m1a is the force on m3.
If it helps please approve.
 

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