### Problem DescriptionWolverine has an arm that is 1 meter long and claws that are 0.5 meters long. He’s attempting to knock over a wall using these dimensions. His shoulder can exert a torque of 1,500 N·m. The task is to determine the ratio of force on the wall with claws extended to the force on the wall without claws extended.### Equations\[\frac{F_{\text{with claws}}}{F_{\text{without claws}}}\]### Answer ChoicesA. 2/3 B. 3/2 C. 1 D. I don't know but the torque direction is ⊗ E. I don't know but the torque direction is ⊙### Diagrams Explanation1. **Wolverine with Claws**: - Arm plus claws total length: 1.5 meters. - Exerted force direction is perpendicular to the Adamantium Wall. - Labeled \( F_{\text{with}} \).2. **Wolverine without Claws**: - Arm length: 1.0 meter. - Exerted force direction is perpendicular to the Adamantium Wall. - Labeled \( F_{\text{without}} \).### Solution Explanation- **Torque Equation**: Torque (\(\tau\)) is given by the product of force (F) and the distance (r) from the pivot point, i.e., \(\tau = F \times r\).- **With Claws**: Torque is 1,500 N·m. Distance (r) = 1.5 m. \( \tau = F_{\text{with}} \times 1.5 \)- **Without Claws**: Torque is 1,500 N·m. Distance (r) = 1.0 m. \( \tau = F_{\text{without}} \times 1.0 \)### Calculation1. Solving for \( F_{\text{with}} \): \[ 1,500 = F_{\text{with}} \times 1.5 \Rightarrow F_{\text{with}} = \frac{1,500}{1.5} = 1,000 \, \text{N} \]2. Solving for \( F_{\text{without}} \):

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Wolverine has an arm that is 1[m] long and claws that are 0.5[m] long. He's trying to knock over a wall. His shoulder can exert a torque of 1,500[N-m]. What is the ratio of force on the wall with vs without his claws extended? 1.5[m] point Pivot F with claws without claws Adamantium Wall А. 2/3 F Wolverine with В. 3/2 with claws С. 1 1.0[m] Pivot Adamantium D. I don't know but the torque direction is point Wall E. I don't know but the torque direction is Fwithout Wolverine Bare hands

Wolverine has an arm that is 1[m] long and claws that are 0.5[m] long. He's trying to knock over a wall. His shoulder can exert a torque of 1,500[N-m]. What is the ratio of force on the wall with vs without his claws extended? 1.5[m] point Pivot F with claws without claws Adamantium Wall А. 2/3 F Wolverine with В. 3/2 with claws С. 1 1.0[m] Pivot Adamantium D. I don't know but the torque direction is point Wall E. I don't know but the torque direction is Fwithout Wolverine Bare hands

College Physics

11th Edition

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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Concept explainers

Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

Question

### Problem Description

Wolverine has an arm that is 1 meter long and claws that are 0.5 meters long. He’s attempting to knock over a wall using these dimensions. His shoulder can exert a torque of 1,500 N·m. The task is to determine the ratio of force on the wall with claws extended to the force on the wall without claws extended.

### Equations

\[
\frac{F_{\text{with claws}}}{F_{\text{without claws}}}
\]

### Answer Choices

A. 2/3
B. 3/2
C. 1
D. I don't know but the torque direction is ⊗
E. I don't know but the torque direction is ⊙

### Diagrams Explanation

1. **Wolverine with Claws**:
- Arm plus claws total length: 1.5 meters.
- Exerted force direction is perpendicular to the Adamantium Wall.
- Labeled \( F_{\text{with}} \).

2. **Wolverine without Claws**:
- Arm length: 1.0 meter.
- Exerted force direction is perpendicular to the Adamantium Wall.
- Labeled \( F_{\text{without}} \).

### Solution Explanation

- **Torque Equation**: Torque (\(\tau\)) is given by the product of force (F) and the distance (r) from the pivot point, i.e., \(\tau = F \times r\).

- **With Claws**: Torque is 1,500 N·m. Distance (r) = 1.5 m.
\( \tau = F_{\text{with}} \times 1.5 \)

- **Without Claws**: Torque is 1,500 N·m. Distance (r) = 1.0 m.
\( \tau = F_{\text{without}} \times 1.0 \)

### Calculation

1. Solving for \( F_{\text{with}} \):
\[
1,500 = F_{\text{with}} \times 1.5 \Rightarrow F_{\text{with}} = \frac{1,500}{1.5} = 1,000 \, \text{N}
\]

2. Solving for \( F_{\text{without}} \):

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