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Chapter 10: Problem 12

A \(15 \mathrm{~N}\) force acts at \(35^{\circ}\) to the \(x\) axis. Resolve the force into forces in the \(x\) and \(y\) directions.

Short Answer

Expert verified
Answer: The x-axis component of the force is approximately 12.29 N, and the y-axis component is approximately 8.60 N.

Step by step solution

01

Identify known values

We are given the following information: - Force magnitude = \(15 \mathrm{~N}\) - Angle with the x-axis = \(35^\circ\)
02

Resolve force into x-axis component

To resolve the force into its x-axis component, we can take the horizontal component of the force. This can be done using the cosine of the angle with the x-axis. The x-axis component of the force \(F_x\) can be calculated as: \(F_x = F \cdot \cos{\theta}\) where \(F\) is the force magnitude, and \(\theta\) is the angle with the x-axis. Plugging in the given values, we get: \(F_x = 15 \cdot \cos{35^\circ}\) \(F_x = 15 \cdot 0.8192\) \(F_x \approx 12.29 \mathrm{~N}\)
03

Resolve force into y-axis component

To resolve the force into the y-axis component, we can take the vertical component of the force. This can be done using the sine of the angle with the x-axis. The y-axis component of the force \(F_y\) can be calculated as: \(F_y = F \cdot \sin{\theta}\) where \(F\) is the force magnitude, and \(\theta\) is the angle with the x-axis. Plugging in the given values, we get: \(F_y = 15 \cdot \sin{35^\circ}\) \(F_y = 15 \cdot 0.5736\) \(F_y \approx 8.60 \mathrm{~N}\)
04

Summarize the results

The force \({15\ \mathrm{N}}\) acting at an angle of \({35^\circ}\) to the x-axis can be resolved into the following components: - x-axis component: \(F_x \approx 12.29\ \mathrm{N}\) - y-axis component: \(F_y \approx 8.60\ \mathrm{N}\)

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Most popular questions from this chapter

In questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{AB}=21 \mathrm{~cm}, \mathrm{AC}=30 \mathrm{~cm}, A=42^{\circ}\)

A point \(\mathrm{A}\) has a bearing of \(45^{\circ}\) and is \(10 \mathrm{~km}\) distant from O. A point B has a bearing of \(160^{\circ}\) and is \(20 \mathrm{~km}\) distant from \(\mathrm{O}\). A point \(\mathrm{C}\) is mid-way between A and B. Calculate (a) the bearing of \(\mathrm{C}\) from \(\mathrm{O}\) (b) the distance of \(\mathrm{C}\) from \(\mathrm{O}\).

In questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(\mathrm{BC}=17 \mathrm{~cm}, \mathrm{AC}=27 \mathrm{~cm}, C=45^{\circ}\)

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(A=36^{\circ}, B=79^{\circ}, \mathrm{AC}=11.63 \mathrm{~cm}\)

For questions \(1-10\) solve \(\triangle \mathrm{ABC}\) given \(B=21^{\circ}, C=46^{\circ}, \mathrm{AB}=9 \mathrm{~cm}\)
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