Question

A bosun's chair is a device used by a boatswain to lift himself to the top of the mainsail of a ship. A simplified device consists of a chair, a rope of negligible mass, and a frictionless pulley attached to the top of the mainsail. The rope goes over the pulley, with one end attached to the chair, and the boatswain pulls on the other end, lifting himself upward. The chair and boatswain have a total mass $M=90.0\mathrm{kg}$. a) If the boatswain is pulling himself up at a constant speed, with what magnitude of force must he pull on the rope? b) If, instead, the boatswain moves in a jerky fashion, accelerating upward with a maximum acceleration of magnitude $a=2.0\mathrm{m}/{\mathrm{s}}^2,$ with what maximum magnitude of force must he null on the rone?

Short answer

Answer: In Scenario 1, the boatswain must pull with a force of 882 N to maintain a constant speed. In Scenario 2, he must pull with a maximum force of 1062 N to achieve a maximum acceleration of 2 m/s² while going upward.

Step-by-step solution

Scenario 1 - Constant Speed

In the first scenario, the boatswain pulls himself up at a constant speed. We must find the magnitude of the force exerted on the rope. Since the speed is constant, the net force on the boatswain-chair system is zero. We can denote the gravitational force as $F_g=Mg$ and the pulling force as $F_p$. According to Newton's second law, the net force on the system is given by $F_{net}=F_p-F_g=0$. Now, we can solve for the pulling force, $F_p$, using the given values for mass and gravitational acceleration (g = 9.8 m/s²): $F_p=F_g=Mg$

Calculation of Force in Scenario 1

We can now plug in the values for mass and gravitational acceleration to find the pulling force: $F_p=(90.0\mathrm{kg})(9.8\mathrm{m}/{\mathrm{s}}^2)$ $F_p=882.0\mathrm{N}$ So in the first scenario, the boatswain must pull with a force of 882 N to maintain a constant speed while going upward.

Scenario 2 - Maximum Acceleration

In the second scenario, the boatswain moves upward with a maximum acceleration of 2 m/s². We now have to find the maximum magnitude of force exerted on the rope. We can use Newton's second law again to find the net force on the system. This time, however, the net force is not zero because the boatswain is accelerating. The net force can be expressed as $F_{net}=F_p-F_g=Ma$, where $M$ is the mass of the boatswain-chair system and $a$ is the acceleration. Now we can solve for $F_p$ in terms of $M$, $g$, and $a$: $F_p=Mg+Ma$

Calculation of Force in Scenario 2

We can now plug in the values for mass, gravitational acceleration, and maximum acceleration: $F_p=(90.0\mathrm{kg})(9.8\mathrm{m}/{\mathrm{s}}^2)+(90.0\mathrm{kg})(2.0\mathrm{m}/{\mathrm{s}}^2)$ $F_p=882.0\mathrm{N}+180.0\mathrm{N}=1062.0\mathrm{N}$ So in the second scenario, the boatswain must pull with a maximum force of 1062 N to achieve a maximum acceleration of 2 m/s² while going upward.

Key concepts

Newton's Second Law

Understanding Newton's second law is vital for solving many physics problems, especially those involving forces and motion. This law can be summed up by the equation,

$$\begin{array}{r}{F}_{net}=ma,\end{array}$$
where $F_{net}$ is the net force acting on an object, $m$ is the mass, and $a$ is the acceleration of the object. This relationship tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

To solve problems using this law, we identify all the forces acting on the object, then sum them to find the net force. Next, we can either find the acceleration if force and mass are known, or solve for the force if mass and acceleration are given. In the bosun's chair problem, when the boatswain is moving at constant speed, the net force is zero, because the gravitational force is perfectly balanced by the pulling force.

However, when the boatswain starts to accelerate, there's an unbalanced force acting on the system, which means the pulling force must increase to not only balance the gravitational force but also provide the additional force required to accelerate.

Gravitational Force

The gravitational force is another cornerstone of physics, particularly when dealing with objects on or near Earth. This force, which attracts two bodies with mass to one another, is represented by the equation,

$$\begin{array}{r}{F}_g=mg,\end{array}$$
where $F_g$ denotes the gravitational force, $m$ is the mass of the object, and $g$ is the acceleration due to gravity, which has an approximate value of $9.8\mathrm{m}/{\mathrm{s}}^2$ near the Earth's surface. The gravitational force plays a huge role in our bosun's chair exercise, where it acts downwards, drawing the boatswain towards the center of the Earth.

In our scenario, when the boatswain pulls on the rope with a force equal to gravitational force, the net force becomes zero, resulting in no acceleration, consistent with moving at a constant speed. It's worth noting that even though the boatswain is not accelerating, he still has to exert a force to counteract the ever-present pull of gravity.

Acceleration

Finally, let's delve into the concept of acceleration. Acceleration is defined as the rate at which an object changes its velocity. It's a vector, meaning it has both a magnitude and a direction. It can be expressed as,

$$\begin{array}{r}a=\frac{\text{change in velocity}}{\text{time taken}},\end{array}$$
In the case of our bosun's chair problem, we distinguish between two forms of motion: constant speed (zero acceleration) and an increasing speed (positive acceleration).

When the boatswain starts accelerating upwards, the force exerted must now overcome not just gravity but also provide enough extra force to accelerate the mass upwards at $2.0\mathrm{m}/{\mathrm{s}}^2$. This changes the dynamics of the problem, requiring the application of Newton's second law to calculate the new required pulling force. Acceleration is crucial in our understanding of how objects move and interact under various forces.