Question

A metal bar is in the $x y$ -plane with one end of the bar at the origin. A force $\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}$ is applied to the bar at the point $x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m} .$ (a) In terms of unit vectors $\hat{\imath}$ and $\hat{\jmath},$ what is the position vector $\vec{r}$ for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by $\overrightarrow{\boldsymbol{F}} ?$

Name: Problem 6, Chapter 10: University Physics with Modern Physics
Uploaded: 2023-10-04T07:57:15-08:00
Duration: 1 min 46 s
Channel: Prabhu Ramji
Description: A metal bar is in the x y -plane with one end of the bar at the origin. A force F=(7.00 N) +(-3.00 N) is applied to the bar at the point x=3.00 m, y=4.00 m . (a) In terms of unit vectors and , what is the position vector r⃗ for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by F ?

   A metal bar is in the $x y$ -plane with one end of the bar at the origin. A force $\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}$ is applied to the bar at the point $x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m} .$ (a) In terms of unit vectors $\hat{\imath}$ and $\hat{\jmath},$ what is the position vector $\vec{r}$ for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by $\overrightarrow{\boldsymbol{F}} ?$

University Physics with Modern Physics

Hugh D. Young 14th Edition

Chapter 10, Problem 6

Instant Answer

Solved by Expert Prabhu Ramji

10/04/2023

Step 1

The position vector is simply a vector that points from the origin to the point in question. In this case, the point is at $x=3.00 \, m$ and $y=4.00 \, m$. Therefore, the position vector is $\vec{r}=(3.00 \, m) \hat{i}+(4.00 \, m) \hat{j}$. Show more…

Show all steps

Thanks for your feedback!

A metal bar is in the $x y$ -plane with one end of the bar at the origin. A force $\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}$ is applied to the bar at the point $x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m} .$ (a) In terms of unit vectors $\hat{\imath}$ and $\hat{\jmath},$ what is the position vector $\vec{r}$ for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by $\overrightarrow{\boldsymbol{F}} ?$