Why Thomas, you're all wet!

We can simplify the situation by taking out the boat and replacing the mast, boom and the sailor by two rods and a point mass respectively (as the question allows us to do/recommends).

In this case, then, the fulcrum/pivot will be the point at which the boat touches the water/ the meeting point of the mast and the boom.

Since the mast is a uniform rod (density is uniform throughout the rods structure), we can assume its center of mass to be at its center, i.e. its weight can be assumed to act at its center which is 14/2 = 7 m from the fulcrum. Thus, the moment of this force (the weight of the mast about the fulcrum along its length) is:

$7 \times 400 \times 9.81 \times cos \theta$

(7 is the length in meters of the line along which the force acts, $400 \times 9.81$ is the weight of the mast(the force) and the $cos \theta$ takes the component of the weight perpendicular to the line of action of the force, i.e. perpendicular to the mast itself)

The moment of the sailor about the boom is then:

$4 \times 100 \times 9.81 \times sin \theta$

(4 m is the length of the line of action of the force about the pivot and $sin \theta$ takes the component of the weight perpendicular to the line of action of this force)

The forces (weights) both act downwards so we need to take their components perpendicular to their respective lines of action (along the mast and boom).

When their moments are equal, any further decrease in $\theta$ will capsize the boat/ unbalance the moments so the maximum value of $\theta$ for which the boat will not capsize is given by

$7 \times 400 \times 9.81 \times cos \theta = 4 \times 100 \times 9.81 \times sin \theta$

$\Rightarrow 7 cos \theta = sin \theta$

$\Rightarrow tan \theta = 7$

Thus, solving this equation gives us

$\theta = 81.87^\circle$

Clearly when the boat is upright ,the torque about the base point due to the weight of the rod (acting in anticlockwise direction in fig) must balance the torque due to the weight of the sailor(acting clockwise here).Now the weight of rod is concentrated about its center of mass and its perpendicular distance about the point of toppling i.e unbalance = 7cos(theta). Torque = |FxD|.Therefore torque due to the weight of rod = 2800cos(theta) Kg.m and torque due to weight of man =400 kg.m. Equating the two torques gives (theta)=cos^-1(1/7) =81.79 degrees.Hence once theta reaches this value,the sailor will be unable to keep the boat upright.

Here we have, Mass of sailor, m1 =100 kg Distance of sailor from point of rotation, l1 =4 m Mass of mast, m2 =400 kg Distance of mast from the point of rotation, l2 =7 m [since the mass is distributed along the whole rod so
The effective distance will be half the main length]

Now, m1l1 sinθ =m2l2 cosθ [Initially the boat was erect but while sailing the boat, it
inclines at angle θ. Now if we consider the will balance their moment horizontal direction, the effective length becomes l2*cosθ. Similarly in the case of sailor ]

Or, θ= tan-1((m1l1)/(m2l2)) Putting the values, we get, θ=81.87

                                                                                                                                                          [solved]

Choose the pivot as the bottom end of the mast. The moments around this pivot are:

1. Counter-clockwise moment of weight of mast: $\tau_{ccw} = m_{mast}gd_{1\perp} = m_{mast}g(\frac {l_1}{2}\cos\theta)$

2. Clockwise moment of weight of sailor: $\tau_{cw} = m_{sailor}gd_{2\perp} = m_{sailor}g({l_2}\sin\theta)$

Principle of Moments: $\tau_{ccw} = \tau_{cw}\\ \Rightarrow m_{mast}g(\frac {l_1}{2}\cos\theta) = m_{sailor}g({l_2}\sin\theta)\\ \Rightarrow \tan\theta = \frac {m_{mast} {l_1}\div{2}} {m_{sailor}l_2}\\ =\frac{400\times14\div2}{100\times4} = 7\\ \Rightarrow \theta = \arctan(7) = 81.9$

as we are considering mast and sail as a uniform rod of length 14 m , we can assume its mass to at 7 m from the boat (center of mass) and thus torque along the boat will be $T_1 = 7*400*g*\cos \theta$ .

And torque due to the sailor will be $T_2 = 4*100*g*\sin \theta$ .

and as sailor is unable to keep the boat upright even if he is sitting at the end of the boom thus $T_1 \geq T_2 ....(1)$

by solving $(1)$ we will get $7 \geq \tan \theta$

therefore $\theta = 81.87$ degrees

At the critical angle where the boat is just about to tip over, the torque of the sailor and the torque of the mast are equal. We therefore have

$\tau_{Mast}=\frac{1}{2} m_{mast}gL_{mast}\cos \theta=m_{sailor}g l_{sailor} \cos (90-\theta)=\tau_{sailor}$ .

This allows us to solve for $\tan \theta=7$ which means $\theta=81.87$ degrees.

The sailboat tipping over is a consequence of angular velocity. The mast's weight and the sailor's weight apply torques to the boat.

Assume that we are at the limit of the sailor's capabilities-- that is, he is sitting at the end of the boom, yet his torque is completely balanced by the mast's torque. Then the angular acceleration will be zero and the boat will continue to tip despite the sailor's efforts.

$T_{tot} = T_{mast} + T_{sailor} = 0$

$T_{sailor} = - (100) g \sin \theta$

Note the negative sign to denote the clockwise direction of the sailor's torque.

The mass of the mast and sails is considered to be distributed uniformly along a 14 meter rod. Because of this, for purposes of calculating torque, we can take the entire mass to be halfway along the rod, that is, at a distance of 7 meters from axis of rotation of the boat.

If you doubt this, consider integrating the torque from infinitesimal bits of weight.

$T = \int_0^L r (\cos \theta) (gM/L) dr$ $= (L^2/2) (\cos \theta) (gM/L) = (gML/2) \cos \theta$

Then the torque from the mast and sails is:

$T_{mast} = g \cdot 400 \cdot 7 \cos \theta$

Taking the torque balance from the beginning:

$-T_{sailor} = T_{mast}$

$400 g \sin \theta = 400 \cdot 7 g \cos \theta$

Solve for the angle:

$\theta = \tan^{-1} (7) = 81.87^\circ$

These solutions depends on the simplifying assumption that the buoyancy of the sailboat, which depends on the cross section of the hull, is not a factor, i.e., the boat might as well be on dry land perched on its keel.