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Chapter 11: Problem 74

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

Expert verified
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

01

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
02

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 \(F_2\)
03

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 \(F_1\)
04

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
05

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
06

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).

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