Question

A construction supervisor of mass \(M=92.1 \mathrm{~kg}\) is standing on a board of mass \(m=27.5 \mathrm{~kg} .\) Two sawhorses at a distance \(\ell=3.70 \mathrm{~m}\) apart support the board, which overhangs each sawhorse by an equal amount. If the man stands a distance \(x_{1}=1.07 \mathrm{~m}\) away from the left-hand sawhorse as shown in the figure, what is the force that the board exerts on that sawhorse?

Short answer

Answer: To find the force exerted by the board on the left-hand sawhorse, follow these steps: 1. Calculate the total weight and position of the center of mass of the system (supervisor and board). 2. Choose a pivot point and list all the torques around that point. In this case, choose the left-hand sawhorse as the pivot point. 3. Set the counterclockwise torque equal to the clockwise torque for the system in static equilibrium. 4. Solve for the force exerted by the right-hand sawhorse, F2. 5. Calculate the force exerted by the left-hand sawhorse, F1, by subtracting F2 from the total weight of the system. Following these steps, you can find the force exerted by the board on the left-hand sawhorse.

Step-by-step solution

Calculate the total weight and position of the center of mass

First, we need to find the total weight of the system (supervisor and board) and its position. This will give us the downward force acting on the sawhorses. The total weight of the system is the sum of the weights of the supervisor and the board: \(W_T= Mg + mg = (92.1 kg)(9.81 \frac{m}{s^2}) + (27.5 kg)(9.81 \frac{m}{s^2})\) Now let's find the position of the center of mass of the system (the combined center of mass of the supervisor and the board). Let's consider the position of the supervisor as \(x_1 = 1.07 m\) and the position of the center of mass of the board as \(x_2 = \frac{\ell}{2} = 1.85 m\). To find the position of the combined center of mass, we can use the following formula: \(x_{cm} = \frac{M x_1 + m x_2}{M + m}\)

Choose a pivot point and list all the torques around that point

Let's choose the left-hand sawhorse as the pivot point (the point around which we calculate the torques). This way, we only need to consider the torques due to the weight of the supervisor and the weight of the board to find the counterclockwise torque around the pivot point \(\tau_{ccw}\): \(\tau_{ccw} = Mg (x_1 - 0) + mg (\frac{\ell}{2} - 0)\) Now let's consider the force exerted by the right-hand sawhorse, \(F_2\). Since the right-hand sawhorse is located \(\ell = 3.7 m\) away from the left-hand sawhorse, the torque due to this force acting clockwise around the pivot point \(\tau_{cw}\) is: \(\tau_{cw} = F_2 (\ell - 0)\)

Set the counterclockwise torque equal to the clockwise torque

Due to the system being in static equilibrium, the counterclockwise and clockwise torques must be equal, so: \(\tau_{ccw} = \tau_{cw}\) Therefore, \(Mg (x_1 - 0) + mg (\frac{\ell}{2} - 0) = F_2 (\ell - 0)\)

Solve for the force exerted by the right-hand sawhorse, F2

We can solve the torque equation from step 3 for the force exerted by the right-hand sawhorse, \(F_2\): \(F_2 = \frac{Mg (x_1 - 0) + mg (\frac{\ell}{2} - 0)}{\ell - 0}\)

Calculate the force exerted by the left-hand sawhorse, F1

Since the system is in equilibrium, the upward forces due to both sawhorses must sum up to the total weight of the system: \(F_1 + F_2= W_T\) We can solve for the force exerted by the left-hand sawhorse, \(F_1\): \(F_1 = W_T - F_2\) Now, we can calculate \(F_1\) using the values from step 1 and step 4. This will give us the force that the board exerts on the left-hand sawhorse.

Key concepts

Torque

Torque is a measure of how much a force acting on an object causes that object to rotate. Understanding torque is crucial for solving problems involving rotational motion and static equilibrium, where objects are not moving. The torque (\tau) can be calculated by the equation \( \tau = F \times d \), where \( F \) is the force applied and \( d \) is the distance from the pivot point to the point where the force is applied, often referred to as the 'lever arm'. Torque has both magnitude and direction; torques that would cause a clockwise rotation are typically considered as negative, while those causing counterclockwise rotation are positive.

The scenario from the exercise showcases how torque is used in real-life situations. A construction supervisor standing on a board creates a rotational effect at the contact points with the sawhorses. By using the distance from the supervisor to the pivot point (the left-hand sawhorse), we calculate his contribution to the torque. Similarly, the center of mass of the board also contributes to the overall torque on the system. Equilibrium is reached when the sum of the torques is zero, meaning no net rotational motion occurs, which is essential in ensuring the board does not tip over. This concept illustrates the importance of torque in static equilibrium scenarios, such as in the construction and stability of structures.

Center of Mass

The center of mass is the point at which the mass distribution of an object is balanced in all directions, essentially where it can be considered to concentrate. For simple objects with uniform density, the center of mass is at the geometric center. However, for composite objects, such as the system in our exercise involving a supervisor and a board, the center of mass depends on the masses and positions of the individual components.

In our problem, we calculate the center of mass of the system to understand how it is shared between the two support points (the sawhorses). A common misconception is that the center of mass must lie within the physical bounds of the object, which isn't always true, as in cases with irregular shapes or where the object's parts are in different locations. In static equilibrium problems, the position of the center of mass helps in determining how forces are distributed across support structures or points of contact. A balanced center of mass is key to ensuring that structures don't tip over, which is crucial in architectural design and object manipulation. Our problem provides an excellent example of how these principles are applied in engineering and physics. When the center of mass is properly understood, it becomes easier to solve for the forces in equilibrium scenarios.

Equilibrium Calculations

Equilibrium calculations are the mathematical process of solving for unknown forces or moments in a system that is in static equilibrium. A system is said to be in static equilibrium when all the forces and torques acting on it are balanced, meaning there is no net force or net torque causing motion.

In problems like our example with the supervisor on a board, equilibrium calculations involve first determining the total weight of the system and then using the conditions for equilibrium: the sum of vertical forces must equal zero, and the sum of torques must also equal zero. Using these conditions allows us to set up equations to solve for unknowns, such as the individual forces exerted by the sawhorses in our scenario. It's crucial to choose a pivot point intelligently, as this can simplify the calculations significantly by eliminating certain unknowns that have no torque contribution about that pivot point. The problem presented requires an intricate understanding of how forces balance out and how torques can provide insights into force distribution. By mastering equilibrium calculations, students can apply these skills to a range of practical problems in engineering, physics, and beyond, from determining the capacity of a shelf to ensuring the stability of a crane.