Tweedledum and Tweedledee are carrying a uniform wooden board that is LLL = 3.00 mm long and has a weight of w = 160 N. Tweedledum applies an upward force of magnitude F1 = 60.0 NN at the left end of the board.Which of the following is the correct equation for Στ for the given rotation axis? All rotation axes lie on the x-axis and torques that would cause a counterclockwise rotation are positive.Enter the letter(s) of the correct equation(s) in alphabetical order. For instance, if you thought that choices A, B, and D were correct, you would enter ABD.Answer Choices: Axis of RotationEtAx=0(1/2)wL+F2d=0Bx=L(1/2)wL-F2(L-d)-F1L=0Cx=0(-1/2)wL+F2d=0Dx=d-F1d-F2(d-(1/2)L)=0Ex=(1/2)L(-1/2)F1L+F2(d-(1/2)L)Fx=(1/2)L(1/2)F1L+F2(d-(1/2)L) **Part D**Which of the following is the correct equation for \( \Sigma \tau \) for the given rotation axis? All rotation axes lie on the x-axis, and torques that would cause a counterclockwise rotation are positive.| Axis of Rotation | \( \Sigma \tau \) ||------------------|-------------------|| A: \( x = 0 \) | \( \frac{1}{2} wL + F_2 d = 0 \) || B: \( x = L \) | \( \frac{1}{2} wL - F_2 (L - d) - F_1 L = 0 \) || C: \( x = 0 \) | \( -\frac{1}{2} wL + F_2 d = 0 \) || D: \( x = d \) | \( -F_1 d - F_2 (d - \frac{1}{2} L) = 0 \) || E: \( x = \frac{1}{2} L \) | \( -\frac{1}{2} F_1 L + F_2 (d - \frac{1}{2} L) \) || F: \( x = \frac{1}{2} L \) | \( \frac{1}{2} F_1 L + F_2 (d - \frac{1}{2} L) \) |Enter the letter(s) of the correct equation(s) in alphabetical order. For instance, if you thought that choices A, B, and D were correct, you would enter ABD. **View Available Hint(s)**

Answer: BCE fig: FBD diagram of the system Let Tweedledum and…

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College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Tweedledum and Tweedledee are carrying a uniform wooden board that is LLL = 3.00 mm long and has a weight of w = 160 N. Tweedledum applies an upward force of magnitude F1 = 60.0 NN at the left end of the board.

Which of the following is the correct equation for Στ for the given rotation axis? All rotation axes lie on the x-axis and torques that would cause a counterclockwise rotation are positive.

Enter the letter(s) of the correct equation(s) in alphabetical order. For instance, if you thought that choices A, B, and D were correct, you would enter ABD.

Answer Choices:

	Axis of Rotation	Et
A	x=0	(1/2)wL+F₂d=0
B	x=L	(1/2)wL-F₂(L-d)-F₁L=0
C	x=0	(-1/2)wL+F₂d=0
D	x=d	-F₁d-F₂(d-(1/2)L)=0
E	x=(1/2)L	(-1/2)F₁L+F₂(d-(1/2)L)
F	x=(1/2)L	(1/2)F₁L+F₂(d-(1/2)L)

**Part D**

Which of the following is the correct equation for \( \Sigma \tau \) for the given rotation axis? All rotation axes lie on the x-axis, and torques that would cause a counterclockwise rotation are positive.

| Axis of Rotation | \( \Sigma \tau \) |
|------------------|-------------------|
| A: \( x = 0 \) | \( \frac{1}{2} wL + F_2 d = 0 \) |
| B: \( x = L \) | \( \frac{1}{2} wL - F_2 (L - d) - F_1 L = 0 \) |
| C: \( x = 0 \) | \( -\frac{1}{2} wL + F_2 d = 0 \) |
| D: \( x = d \) | \( -F_1 d - F_2 (d - \frac{1}{2} L) = 0 \) |
| E: \( x = \frac{1}{2} L \) | \( -\frac{1}{2} F_1 L + F_2 (d - \frac{1}{2} L) \) |
| F: \( x = \frac{1}{2} L \) | \( \frac{1}{2} F_1 L + F_2 (d - \frac{1}{2} L) \) |

Enter the letter(s) of the correct equation(s) in alphabetical order. For instance, if you thought that choices A, B, and D were correct, you would enter ABD.

**View Available Hint(s)**

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