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Chapter 3: Problem 344

Three Forces \(F_{1}, F_{2}\), and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(\mathrm{X}\) - axis, \(\mathrm{F}_{2}=4 \mathrm{~N}\) along the positive Y-axis then the third force \(F_{3}\) is (A) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (B) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis (C) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (D) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis

Short Answer

Expert verified
The third force \(F_{3}\) is 5 N, making an angle \(\theta = \tan^{-1}(3/4)\) with the negative Y-axis. Therefore, the correct answer is (A).

Step by step solution

01

Write down the given forces

We are given 3 forces acting on the body: \( F_1 = 3 \, N \) along the positive X-axis, \( F_2 = 4 \, N \) along the positive Y-axis, and \( F_3 \) that will balance out the other two forces.
02

Find the X and Y components of the third force

For the body to be in equilibrium, the net force along both X and Y axis should be zero. So the X and Y components of F3 should cancel out with the given forces. Along X-axis, \(F_{3x} = - F_{1} = -3 \, N\) Along Y-axis, \(F_{3y} = - F_{2} = -4 \, N\)
03

Calculate the magnitude and angle of F3

Now we can find the magnitude of the third force using the Pythagorean theorem: \(F_{3} = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(-3)^2 + (-4)^2} = 5\, N\) We find the angle F3 makes with the negative Y-axis using the tangent function: \(\theta = \tan^{-1}\left(\frac{F_{3x}}{-F_{3y}}\right) = \tan^{-1}\left(\frac{-3}{-4}\right) = \tan^{-1}\left(\frac{3}{4}\right)\) Therefore, the third force F3 is 5 N, making an angle \(\theta = \tan^{-1}(3/4)\) with the negative Y-axis. This corresponds to option (A).

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