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Free body diagram:
Given data:
Length of massless scaffold (end to end) L=12 m.
Weight of box m=300 N.
Length of massless scaffold (center to end) l=6 m.
As, the box sits 4.0 m from the left end, the distance of the box from the center of massless scaffold is given as,
[tex]\begin{gathered} r=6.0\text{ m}-4.0\text{ m} \\ =2.0\text{ m} \end{gathered}[/tex]Balancing force in y direction,
[tex]T_1+T_2=300\text{ N}\ldots(1)[/tex]The torque is given as,
[tex]\tau=perpendicular\text{ distance}\times force[/tex]Therefore, torque along the center of massless scaffold is given as,
[tex]\begin{gathered} \Sigma\tau=0 \\ l\times T_1+r\times(300\text{ N})-l\times\tau_2=0 \\ 6\times T_1+2\times(300\text{ N})-6\times T_2=0 \\ 6T_1+600\text{ N}-6T_2=0 \\ 6(T_1+100\text{ N}-T_2)=0 \\ T_1+100\text{ N}-T_2=0 \\ T_1-T_2=-100\text{ N}\ldots(2) \end{gathered}[/tex]Adding equation (1) and (2),
[tex]\begin{gathered} (T_1+T_2)+(T_1-T_2)=300\text{ N}-100\text{ N} \\ T_1+T_1+T_2-T_2=200\text{ N} \\ 2T_1=200\text{ N} \\ T_1=\frac{200\text{ N}}{2} \\ T_1=100\text{ N} \end{gathered}[/tex]Substituting T1 in equation (1) we get,
[tex]\begin{gathered} 100\text{ N}+T_2=300\text{ N} \\ T_2=300\text{ N}-100\text{ N} \\ T_2=200\text{ N} \end{gathered}[/tex]Therefore, tension in left wire is 100 N and tension on right wire is 200 N. Hence, option (1) is the correct choice.
