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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.
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.SoT2-T1=m2aor,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.SoT3-T2=m1aT3=T2+m1a is the force on m3.If it helps please approve.
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