The force exerted on the diving board at A and B.

Question

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Answer 1

Question

Chapter 9, Problem 1OQ

To determine

The force exerted on the diving board at A and B.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Option (a)F_A is down, F_B is up and F_B is larger than F_A.

Explanation of Solution

Torque can be understood as a force that tends to cause rotation.

$\begin{array}{l} τ = r \times F \\ = r F \sin θ \end{array}$

Weight of the diver will act downwards. The reaction forces will act upwards at A and B.

Physics: Principles and Applications -- Pearson e Text Instant Access (Pearson+), Chapter 9, Problem 1OQ

$\begin{array}{l} τ_{A} = 0 \\ F_{B} (x_{A} - x_{B}) = M g x_{A} \\ \Rightarrow F_{B} = \frac{M g x_{A}}{(x_{A} - x_{B})} ... (1) \\ τ_{B} = 0 \\ F_{A} (x_{A} - x_{B}) = M g x_{B} \\ \Rightarrow F_{A} = \frac{M g x_{B}}{(x_{A} - x_{B})} ... (2) \\ ∵ x_{A} > x_{B} \\ ∴ F_{B} > F_{A} \end{array}$

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Answer 2

Question

Chapter 9, Problem 1OQ

To determine

The force exerted on the diving board at A and B.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Option (a)F_A is down, F_B is up and F_B is larger than F_A.

Explanation of Solution

Torque can be understood as a force that tends to cause rotation.

$\begin{array}{l} τ = r \times F \\ = r F \sin θ \end{array}$

Weight of the diver will act downwards. The reaction forces will act upwards at A and B.

Physics: Principles and Applications -- Pearson e Text Instant Access (Pearson+), Chapter 9, Problem 1OQ

$\begin{array}{l} τ_{A} = 0 \\ F_{B} (x_{A} - x_{B}) = M g x_{A} \\ \Rightarrow F_{B} = \frac{M g x_{A}}{(x_{A} - x_{B})} ... (1) \\ τ_{B} = 0 \\ F_{A} (x_{A} - x_{B}) = M g x_{B} \\ \Rightarrow F_{A} = \frac{M g x_{B}}{(x_{A} - x_{B})} ... (2) \\ ∵ x_{A} > x_{B} \\ ∴ F_{B} > F_{A} \end{array}$

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Answer 3

Question

Chapter 9, Problem 1OQ

To determine

The force exerted on the diving board at A and B.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Option (a)F_A is down, F_B is up and F_B is larger than F_A.

Explanation of Solution

Torque can be understood as a force that tends to cause rotation.

$\begin{array}{l} τ = r \times F \\ = r F \sin θ \end{array}$

Weight of the diver will act downwards. The reaction forces will act upwards at A and B.

Physics: Principles and Applications -- Pearson e Text Instant Access (Pearson+), Chapter 9, Problem 1OQ

$\begin{array}{l} τ_{A} = 0 \\ F_{B} (x_{A} - x_{B}) = M g x_{A} \\ \Rightarrow F_{B} = \frac{M g x_{A}}{(x_{A} - x_{B})} ... (1) \\ τ_{B} = 0 \\ F_{A} (x_{A} - x_{B}) = M g x_{B} \\ \Rightarrow F_{A} = \frac{M g x_{B}}{(x_{A} - x_{B})} ... (2) \\ ∵ x_{A} > x_{B} \\ ∴ F_{B} > F_{A} \end{array}$

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Answer 4

Question

Chapter 9, Problem 1OQ

To determine

The force exerted on the diving board at A and B.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Option (a)F_A is down, F_B is up and F_B is larger than F_A.

Explanation of Solution

Torque can be understood as a force that tends to cause rotation.

$\begin{array}{l} τ = r \times F \\ = r F \sin θ \end{array}$

Weight of the diver will act downwards. The reaction forces will act upwards at A and B.

Physics: Principles and Applications -- Pearson e Text Instant Access (Pearson+), Chapter 9, Problem 1OQ

$\begin{array}{l} τ_{A} = 0 \\ F_{B} (x_{A} - x_{B}) = M g x_{A} \\ \Rightarrow F_{B} = \frac{M g x_{A}}{(x_{A} - x_{B})} ... (1) \\ τ_{B} = 0 \\ F_{A} (x_{A} - x_{B}) = M g x_{B} \\ \Rightarrow F_{A} = \frac{M g x_{B}}{(x_{A} - x_{B})} ... (2) \\ ∵ x_{A} > x_{B} \\ ∴ F_{B} > F_{A} \end{array}$

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 9, Problem 1OQ

To determine

The force exerted on the diving board at A and B.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Option (a)F_A is down, F_B is up and F_B is larger than F_A.

Explanation of Solution

Torque can be understood as a force that tends to cause rotation.

$\begin{array}{l} τ = r \times F \\ = r F \sin θ \end{array}$

Weight of the diver will act downwards. The reaction forces will act upwards at A and B.

Physics: Principles and Applications -- Pearson e Text Instant Access (Pearson+), Chapter 9, Problem 1OQ

$\begin{array}{l} τ_{A} = 0 \\ F_{B} (x_{A} - x_{B}) = M g x_{A} \\ \Rightarrow F_{B} = \frac{M g x_{A}}{(x_{A} - x_{B})} ... (1) \\ τ_{B} = 0 \\ F_{A} (x_{A} - x_{B}) = M g x_{B} \\ \Rightarrow F_{A} = \frac{M g x_{B}}{(x_{A} - x_{B})} ... (2) \\ ∵ x_{A} > x_{B} \\ ∴ F_{B} > F_{A} \end{array}$

Answer 6

Chapter 9, Problem 1OQ

To determine

The force exerted on the diving board at A and B.

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Option (a)F_A is down, F_B is up and F_B is larger than F_A.

Explanation of Solution

Torque can be understood as a force that tends to cause rotation.

$\begin{array}{l} τ = r \times F \\ = r F \sin θ \end{array}$

Weight of the diver will act downwards. The reaction forces will act upwards at A and B.

Physics: Principles and Applications -- Pearson e Text Instant Access (Pearson+), Chapter 9, Problem 1OQ

$\begin{array}{l} τ_{A} = 0 \\ F_{B} (x_{A} - x_{B}) = M g x_{A} \\ \Rightarrow F_{B} = \frac{M g x_{A}}{(x_{A} - x_{B})} ... (1) \\ τ_{B} = 0 \\ F_{A} (x_{A} - x_{B}) = M g x_{B} \\ \Rightarrow F_{A} = \frac{M g x_{B}}{(x_{A} - x_{B})} ... (2) \\ ∵ x_{A} > x_{B} \\ ∴ F_{B} > F_{A} \end{array}$

Answer 7

Chapter 9, Problem 1OQ

To determine

The force exerted on the diving board at A and B.

Answer 8

Expert Solution & Answer

Answer to Problem 1OQ

Solution:

Option (a)F_A is down, F_B is up and F_B is larger than F_A.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Torque can be understood as a force that tends to cause rotation.

$\begin{array}{l} τ = r \times F \\ = r F \sin θ \end{array}$

Weight of the diver will act downwards. The reaction forces will act upwards at A and B.

Physics: Principles and Applications -- Pearson e Text Instant Access (Pearson+), Chapter 9, Problem 1OQ

$\begin{array}{l} τ_{A} = 0 \\ F_{B} (x_{A} - x_{B}) = M g x_{A} \\ \Rightarrow F_{B} = \frac{M g x_{A}}{(x_{A} - x_{B})} ... (1) \\ τ_{B} = 0 \\ F_{A} (x_{A} - x_{B}) = M g x_{B} \\ \Rightarrow F_{A} = \frac{M g x_{B}}{(x_{A} - x_{B})} ... (2) \\ ∵ x_{A} > x_{B} \\ ∴ F_{B} > F_{A} \end{array}$