Question

A uniform beam of length L = 7.50 m and weight 4.70 x io2 N is carried by two workers, Sam and Joe, as shown in the figure below. Determine the force that each person exerts on the beam. N

Answer 1

Let \(F_{3}, F_{\mathrm{J}}\) are the forces acting on pers ons Sam and Joe respectivelu.

The separation between Sam and center of the rod is,

$$ L_{1}=\frac{7.50 \mathrm{~m}}{2}-1.0 \mathrm{~m}=2.75 \mathrm{~m} $$

The separation between Sam and Joe is,

$$ L_{2}=7.50 \mathrm{~m}-1.0 \mathrm{~m}-2.0 \mathrm{~m}=4.5 \mathrm{~m} $$

Apply torque condition about position of Sam

$$ \begin{array}{c} F_{s}(0)-M g\left(L_{1}\right)+F_{j}\left(L_{2}\right)=0 \\ F_{j}=\frac{M g\left(L_{1}\right)}{L_{2}}=\frac{\left(4.7 \times 10^{2} \mathrm{~N}\right)(2.75 \mathrm{~m})}{(4.5 \mathrm{~m})} \\ F_{j}=2.87 \times 10^{2} \mathrm{~N} \end{array} $$

Apply Newton's second law to the system.

$$ \begin{aligned} F_{3}+F_{j} &=M g \\ F_{s} &=M g-F_{j} \\ F_{3} &=4.7 \times 10^{2} \mathrm{~N}-2.87 \times 10^{2} \mathrm{~N} \\ F_{s} &=1.83 \times 10^{2} \mathrm{~N} \end{aligned} $$