It's just a plank, bro.

Classical Mechanics Level 1

A wooden plank of length 1 1 m m and uniform cross-section is hinged at one end to the bottom of a tank as shown in the figure below. The tank is filled with water up to a height of 0.5 0.5 m m . The specific gravity of the plank is 0.5 0.5 . Find the angle θ \theta that the plank makes with the vertical in the equilibrium position such that θ ≠ 0 o , 9 0 o \theta \neq 0^{o}, 90^{o} .

1 5 o 15^{o} 6 0 o 60^{o} 4 5 o 45^{o} 3 0 o 30^{o}

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The forces acting on the plank are shown in the figure. The height of water level is x x = 0.5 m. The length of the plank is 2 x 2x = 1 m.

The weight of the plank acts through the centre C of the plank. We have AC = 0.5 m. The buoyant force B B acts through the point B, which is the middle point of the dipped part AD of the plank.

So, AB = A D 2 \frac{AD}{2} = x 2 cos ⁡ θ \frac{x}{2\cos\theta}

Let the mass per unit length of the plank be p p .

Plank's weight, W W = m g mg = p ( 2 x ) g p(2x)g = 2 p x g 2pxg

The mass of the part AD of the plank = x p cos ⁡ θ \frac{xp}{\cos\theta}

The mass of water displaced = 1 specific gravity \frac{1}{\text{specific gravity}} x p cos ⁡ θ \frac{xp}{\cos\theta} = 2 x p cos ⁡ θ \frac{2xp}{\cos\theta}

So the buoyant force, B B = 2 x p g cos ⁡ θ \frac{2xpg}{\cos\theta}

Now, for equilibrium, the torque of m g mg about C should balance the torque of B B about B.

So, m g mg (AC) sin ⁡ θ \sin\theta = B B (AB) sin ⁡ θ \sin\theta

⟹ \implies 2 p x g × x 2pxg \times x = 2 x p g cos ⁡ θ × x 2 cos ⁡ θ \frac{2xpg}{\cos\theta} \times \frac{x}{2\cos\theta}

⟹ \implies cos ⁡ 2 θ \cos^{2}\theta = 1 2 \frac{1}{2}

⟹ \implies cos ⁡ θ \cos\theta = 1 2 \frac{1}{\sqrt2} = 4 5 o \boxed{45^{o}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...