Question

A 2.00 -m-long diving board of mass \(12.0 \mathrm{~kg}\) is \(3.00 \mathrm{~m}\) above the water. It has two attachments holding it in place. One is located at the very back end of the board, and the other is \(25.0 \mathrm{~cm}\) away from that end. a) Assuming that the board has uniform density, find the forces acting on each attachment (take the downward direction to be positive). b) If a diver of mass \(65.0 \mathrm{~kg}\) is standing on the front end, what are the forces acting on the two attachments?

Short answer

Question: Calculate the forces acting on each attachment of the diving board both when it has no diver on it and when a diver is standing on the front end. Answer: Without a diver, the forces acting at each attachment are \(F_1 = -352.8\ N\) (upwards) and \(F_2 = 470.4\ N\) (downwards). With a diver on the front end, the forces acting at each attachment are \(F_1 = -4811.8\ N\) (upwards) and \(F_2 = 5566.4\ N\) (downwards).

Step-by-step solution

Let \(F_1\) represent the force acting at the first attachment (rear end) and \(F_2\) represent the force acting at the second attachment (25cm from the rear end). Let \(W_b\) be the weight of the diving board and \(W_d\) be the weight of the diver. Step 2: Calculate the weight of the diving board (\(W_b\))

To obtain the weight of the diving board, we will use the mass given in the problem and multiply it by the acceleration due to gravity (approximately \(9.8 m/s^2\)). So, \(W_b = 12.0\ kg \times 9.8\ m/s^2 = 117.6\ N\) Step 3: Calculate the distance from the center of mass

Since the diving board has a uniform density, its center of mass will be at the midpoint, i.e., \(1.00\ m\) from the rear end. Step 4: Set up the torque equation

Using the first attachment as the pivot point, the torques due to \(F_2\) and \(W_b\) must balance for the diving board to be in static equilibrium. Therefore, we can write the torque equation as: \(F_2 \times 0.25\ m = W_b \times 1.00\ m\). Step 5: Solve for \(F_2\)

Dividing both sides of the torque equation by \(0.25\ m\), we get \(F_2 = \frac{W_b \times 1.00\ m}{0.25\ m} = \frac{117.6\ N}{0.25\ m} = 470.4\ N\) Step 6: Set up the force equation

The vertical forces acting on the diving board must balance for it to be in static equilibrium. Therefore, we can write the force equation as: \(F_1 + F_2 = W_b\) Step 7: Solve for \(F_1\)

Plugging in the value of \(F_2\) and \(W_b\), we get \(F_1 = W_b - F_2 = 117.6\ N - 470.4\ N = -352.8\ N\). Since the force is negative, it means \(F_1\) is acting upwards. So, without a diver, the forces acting at each attachment are \(F_1 = -352.8\ N\) (upwards) and \(F_2 = 470.4\ N\) (downwards). b) Finding the forces with the diver: Step 1: Calculate the weight of the diver (\(W_d\))

Likewise, we can obtain the weight of the diver by multiplying his mass by the acceleration due to gravity: \(W_d = 65.0\ kg \times 9.8\ m/s^2 = 637\ N\) Step 2: Rewrite the torque and force equations

Taking into account the weight of the diver, the torque equation becomes: \(F_2 \times 0.25\ m = W_b \times 1.00\ m + W_d \times 2.00\ m\) Step 3: Solve for \(F_2\) with the diver

Substituting the values of \(W_b\) and \(W_d\), we get \(F_2 \times 0.25\ m = 117.6\ N \times 1.00\ m + 637\ N \times 2.00\ m\), which yields \(F_2 = \frac{1391.6\ N}{0.25\ m} = 5566.4\ N\) Step 4: Rewrite the force equation with the diver

The force equation now becomes: \(F_1 + F_2 = W_b + W_d\) Step 5: Solve for \(F_1\) with the diver

Plugging in the values of \(F_2\), \(W_b\), and \(W_d\), we get \(F_1 = (W_b + W_d) - F_2 = (117.6\ N + 637\ N) - 5566.4\ N = -4811.8\ N\). Again, since the force is negative, it means \(F_1\) is acting upwards. So, with a diver on the front end, the forces acting at each attachment are \(F_1 = -4811.8\ N\) (upwards) and \(F_2 = 5566.4\ N\) (downwards).

Key concepts

Torque

Torque is a crucial concept when dealing with rotational motion. Imagine trying to open a door. You naturally apply force at the handle, far from the hinges. This creates more torque or rotational force. Torque is defined mathematically as the product of force and the distance from the pivot point: \( \tau = F \times d \).
If your goal is to rotate an object, such as a diving board, balance of torque assures it doesn't just spin around uncontrollably. In the exercise, when assuming one attachment as the pivot, we calculated the forces to achieve equilibrium in torques. This helps us understand how forces distribute across differing points on the object.

Force equilibrium

In the context of statics, force equilibrium is a state where all the forces acting on a body sum to zero, ensuring it remains static or still. One way to visualize this is if you're holding a book in your hand. The upward force from your hand must balance the downward force of gravity on the book to keep it from falling.
In the diving board problem, two forces, \(F_1\) and \(F_2\), act at the attachments, while the weights of the board and potentially a diver contribute to the total load. For equilibrium, these must all balance: \( F_1 + F_2 = W_b + W_d \). This equation ensures that the board doesn't accelerate in part due to forces canceling each other out.

Center of mass

The center of mass of any homogenous object sits at its geometric center if its density is uniform. This point is where all of an object's mass is considered to be concentrated, and it's the point about which the object can uniformly balance.
In terms of the diving board, the center of mass lies halfway along the length, aiding in calculations concerning equilibrium and torque. With uniform density, assuming the midpoint allows for straightforward estimations about how the board will react to forces and torques applied elsewhere. This is foundational in understanding lever effects and rotational systems.

Uniform density

Objects possessing uniform density share a constant density across their volume. With uniform density, the object's mass distribution is constant throughout, leading to predictable locations for their center of mass.
For example, the diving board's center aligns with its geometric midpoint due to its uniform density. This simplifies calculations in statics, specifically those revolving around torque and center of mass. Knowing one section of the board doesn't weigh more than another helps confidently predict how forces distribute, facilitating straightforward problem-solving.