**Statement:**

A
vessel of constant rectangular cross-section is 300 ft long and 35 ft broad,
and floats at an even keel draught of 10 ft. in S.W. The weight of the vessel
may be taken as being distributed uniformly over its entire length.

Where
should a weight be placed in order to give no bending moment at amidships?

**Solution:**

Additional moment caused by the weight will be equal to the additional buoyancy caused by it.

Now,

`TPI=\frac{300\times35}{420}=25`

MCT 1" = `\frac{∆BML}{12L}` = `\frac{BL^2} {420×12}` = `\frac{35×300^2}{420×12}`=`625`

Let, w= weight placed at a distance aft from amidships

d = parallel sinkage

x’ = trim

Hence,

`d=\frac{w}{25}` in = `\frac{w}{300}` ft

`x^\prime` = `\frac{\text{Moment To Change Trim}}{\text{MCT 1}\prime\prime}\times\frac{l}{L}`

`=\frac{wx\times150}{625\times300}` in

`=\frac{wx}{625\times2\times12}ft`

Now,

`W = B`

`or,\ wx=\frac{d+d+x^\prime}{2}\times\frac{50\times35}{35}\times{75+\frac{150}{6}\times\frac{d+x^\prime-d}{d+x^\prime+d}}`

`=11250d+7500x\prime`

Putting the value of d and x’.

`x` = `75` ft