Question

A uniform beam of length $7.60m$ and weight $4.50\times {10}^{2}N$ is carried by two workers, Sam and Joe, as shown in figure $P12.11$. Determine the forces that each person exerts on the beam.

Answer 1

Given,

Length of beam is $7.6m$

Weight of of the beam is $4.5\times {10}^{2}N$

Here, beam is in equilibrium therefore total torque will be zero, therefore,

$F_{sam}(\frac{7.6}{2}-1)=F_{joe}(\frac{7.6}{2}-2)F_{sam}(3.8-1)=F_{joe}(3.8-2)F_{sam}\left(2.8\right)=F_{joe}\left(1.8\right)F_{sam}=\frac{1.8}{2.8}{F}_{joe}{F}_{sam}=0.64F_{joe}$

And

$F_{sam}+F_{joe}=4.5\times {10}^2$

Substitute force of sam in above equation

$0.64F_{joe}+F_{joe}=4.5\times {10}^2{F}_{joe}=\frac{4.5\times {10}^2}{0.64+1}{F}_{joe}=274.39N$

$F_{sam}=0.64\times 274.39F_{sam}=175.6N$

Hence, force exerted by sam is $175.6N$ and force exerted by joe is $274.39N$