Question

Cantilever beam is subjected to linearly varying load boundary condition

Answer 1

Explanation:

Let F be the foot and S be the summit of the

mountain FOS.

Then, ∠OFS = 45º and therefore ∠OSF = 45º.

Consequently,

OF = OS = h km(say).

Let FP = 1000 m = 1 km be the slope so that

∠OFP = 30º. Draw PM ⊥OS and PL ⊥OF.

Join PS. It is given that ∠MPS = 60º.

In ∆FPL, We have

`\text{sin }30^\text{o}=\frac{PL}{PF}`

`\Rightarrow PL=PF\text{ sin }30^\text{o}=( 1\times

\frac{1}{2})=\frac{1}{2}km.`

`\therefore OM=PL=\frac{1}{2}km`

`\Rightarrow MS=OS-OM=( h-\frac{1}{2})km`

`\text{Also, cos }30^\text{o}=\frac{FL}{PF}`

`\Rightarrow FL=PF\text{ cos }30^{o}=( 1\times\frac

{\sqrt{3}}{2})=\frac{\sqrt{3}}{2}km`

Now, h = OS = OF = OL + LF

`\Rightarrow h=OL+\frac{\sqrt{3}}{2}`

`\Rightarrow OL=( h-\frac{\sqrt{3}}{2})km`

`\Rightarrow PM=( h-\frac{\sqrt{3}}{2})km`

In ∆PSM, we have

`\text{tan }60^\text{o}=\frac{SM}{PM}`

⇒ SM = PM. tan 60º …..(ii)

`\Rightarrow ( h-\frac{1}{2})=( h-\frac{\sqrt{3}}

{2})\sqrt{3}`

[Using equations (i) and (ii)]

`\Rightarrow h-\frac{1}{2}=h\sqrt{3}-\frac{3}{2}`

⇒ h(√3 – 1) = 1

`\Rightarrow h=\frac{1}{\sqrt{3}-1}`

`\Rightarrowh=\frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt

{3}+1)}=\frac{\sqrt{3}+1}{2}`

`=\frac{2.732}{2}=1.336km`

Hence, the height of the mountain is 1.366 km