**Problem Statement:**Three forces act on a particle, but it remains stationary. Two of the forces are \(\vec{F_1} = 12.3\hat{i} - 6.50\hat{j}\) and \(\vec{F_2} = -9.60\hat{i} + 8.70\hat{j}\), both in newtons. What is the magnitude (N) of the third force?**Options:**- 8.89- 1.67- 3.48- 6.77- 10.6- 4.87- 5.68- 11.8- 2.46- 7.81**Explanation:**To solve this problem, we use the fact that the net force on the particle is zero (since it remains stationary). This means the vector sum of all forces acting on the particle is zero. If we denote the third force as \(\vec{F_3}\), then we have:\[\vec{F_1} + \vec{F_2} + \vec{F_3} = \vec{0}\]Thus, \(\vec{F_3} = -(\vec{F_1} + \vec{F_2})\).**Calculations:**First, find the sum of \(\vec{F_1}\) and \(\vec{F_2}\):\[\vec{F_1} + \vec{F_2} = (12.3\hat{i} - 6.50\hat{j}) + (-9.60\hat{i} + 8.70\hat{j}) \\= (12.3 - 9.60)\hat{i} + (-6.50 + 8.70)\hat{j} \\= 2.70\hat{i} + 2.20\hat{j}\]Then, compute \(\vec{F_3}\):\[\vec{F_3} = -(2.70\hat{i} + 2.20\hat{j}) \\= -2.70\hat{i} - 2.20\hat{j}\]Finally, calculate the magnitude of \(\vec{F_3

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Three forces act on a particle, but it remains stationary. Two of the forces are in newtons. What is the magnitude (N) of the third force? = 12.31 -6.50 and F₂ = -9.60 +8.707, both

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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Question

**Problem Statement:**

Three forces act on a particle, but it remains stationary. Two of the forces are \(\vec{F_1} = 12.3\hat{i} - 6.50\hat{j}\) and \(\vec{F_2} = -9.60\hat{i} + 8.70\hat{j}\), both in newtons. What is the magnitude (N) of the third force?

**Options:**

- 8.89
- 1.67
- 3.48
- 6.77
- 10.6
- 4.87
- 5.68
- 11.8
- 2.46
- 7.81

**Explanation:**

To solve this problem, we use the fact that the net force on the particle is zero (since it remains stationary). This means the vector sum of all forces acting on the particle is zero. If we denote the third force as \(\vec{F_3}\), then we have:

\[
\vec{F_1} + \vec{F_2} + \vec{F_3} = \vec{0}
\]

Thus, \(\vec{F_3} = -(\vec{F_1} + \vec{F_2})\).

**Calculations:**

First, find the sum of \(\vec{F_1}\) and \(\vec{F_2}\):

\[
\vec{F_1} + \vec{F_2} = (12.3\hat{i} - 6.50\hat{j}) + (-9.60\hat{i} + 8.70\hat{j}) \\
= (12.3 - 9.60)\hat{i} + (-6.50 + 8.70)\hat{j} \\
= 2.70\hat{i} + 2.20\hat{j}
\]

Then, compute \(\vec{F_3}\):

\[
\vec{F_3} = -(2.70\hat{i} + 2.20\hat{j}) \\
= -2.70\hat{i} - 2.20\hat{j}
\]

Finally, calculate the magnitude of \(\vec{F_3

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