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A smooth massless string passes over a smooth fixed pulley. Two masses m1 and m_(2) (m_(1) gt m_(2)) are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is |
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Answer» `(m_(1)+m_(2))g`  Let a be the common acceleration of the system and T be the tension in the string. The equations of motion of two masses are `m_(1)g-T=m_(1)a` …(i) `T-m_(2)g=m_(2)a` …(II) Adding eqn. (i) and eqn. (ii), we get `a=((m_(1)-m_(2)))/(m_(1)+m_(2))g` ...(iii) Acceleration of centre of mass of the system is `a_(CM)=(m_(1)a_(1)+m_(2)a_(2))/(m_(1)+m_(2))=(m_(1)a-m_(2)a)/(m_(1)+m_(2))` (`becausea_(1)` and `a_(2)` are equal in magnitude but opposite in direction) `a_(CM)=((m_(1)-m_(2))/(m_(1)-m_(2)))a=((m_(1)-m_(2))/(m_(1)+m_(2)))((m_(1)-m_(2))/(m_(1)+m_(2)))g` (Using (iii)) `=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g` The TOTAL external force ACTING on the two masses is `F_("ext")=(m_(1)+m_(2))a_(CM)=(m_(1)+m_(2))((m_(1)+m_(2))/(m_(1)+m_(2)))^(2)g` `=((m_(1)-m_(2))^(2))/(m_(1)+m_(2))g` |
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