A force \( \mathbf{F}=[7,6] \) is applied to move a car in the direction \( \mathbf{d}=[-8,-9] \). What is the magnitude of the force in the direction of motion? magnitude of \( \mathbf{F}= \) Ex. 1.2
Upstudy AI Solution
Answer
Solution

Sign in to Unlock Answers for Free!
A Learning Platform Trusted by Millions of Real Students and Teachers.

Answered by UpStudy AI and reviewed by a Professional Tutor
Extra Insights
To find the magnitude of the force in the direction of motion, we first need to find the unit vector of the direction. The magnitude of the direction vector \( \mathbf{d} = [-8, -9] \) is calculated as \( \sqrt{(-8)^2 + (-9)^2} = \sqrt{64 + 81} = \sqrt{145} \). The unit vector \( \mathbf{u_d} \) in the direction of \( \mathbf{d} \) is then \( \mathbf{u_d} = \frac{1}{\sqrt{145}} [-8, -9] = \left[-\frac{8}{\sqrt{145}}, -\frac{9}{\sqrt{145}}\right] \). The magnitude of the force in the direction of motion can be found using the dot product: \[ F_{motion} = \mathbf{F} \cdot \mathbf{u_d} = [7, 6] \cdot \left[-\frac{8}{\sqrt{145}}, -\frac{9}{\sqrt{145}}\right] = -\frac{56}{\sqrt{145}} - \frac{54}{\sqrt{145}} = -\frac{110}{\sqrt{145}}. \] The magnitude we are looking for is the absolute value \( \left| F_{motion} \right| = \frac{110}{\sqrt{145}} \). So the final answer for the magnitude of the force in the direction of motion is approximately \( F_{motion} \approx 9.16 \).