Skip to main content
Physics

Return to Question

added 3 characters in body; edited tags; edited title
Source Link
Qmechanic ♦
Qmechanic ♦
  • 212.9k
  • 48
  • 589
  • 2.3k

Why is the force applied at end of rod by the axis of rotation (R)$R$, at time, $t=0$, equal to $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$?

the question is regarding a problem in Rotational Motion for rigid bodies in AP Physics 1. I have tried to translate it as best as I could from Swedish. The problem is as follows: " A homogeneous rod with length L and mass M is fixed to the horizontal axis. The rod can freely move around its axis. Initially, the rod is standing vertically. At t=0, a force F is applied at the end of the Rod (see image)

What is the magnitude of the Reaction Force, R, that the axis of rotation applies to the rod at t=0? "

"A homogeneous rod with length $L$ and mass $M$ is fixed to the horizontal axis. The rod can freely move around its axis. Initially, the rod is standing vertically. At t=0, a force F is applied at the end of the Rod (see image)

What is the magnitude of the Reaction Force, R, that the axis of rotation applies to the rod at t=0?"

An attempt to draw the Free-Body Diagram

I have tried to draw a Free-Body Diagram with the "Reaction Force" split into its x- and y-components. My teacher has published the solution, which is $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$ with no explanation... I understand that we are finding the hypothenuse R via Pythagoras but it is especially the fact that the solution implies that $Rx = \frac{1}{2} \cdot F$. I cant seem to wrap my head around as to why though? And how do we know so?

Why is the force applied at end of rod by the axis of rotation (R), at time, $t=0$, equal to $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$?

the question is regarding a problem in Rotational Motion for rigid bodies in AP Physics 1. I have tried to translate it as best as I could from Swedish. The problem is as follows: " A homogeneous rod with length L and mass M is fixed to the horizontal axis. The rod can freely move around its axis. Initially, the rod is standing vertically. At t=0, a force F is applied at the end of the Rod (see image)

What is the magnitude of the Reaction Force, R, that the axis of rotation applies to the rod at t=0? "

An attempt to draw the Free-Body Diagram

I have tried to draw a Free-Body Diagram with the "Reaction Force" split into its x- and y-components. My teacher has published the solution, which is $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$ with no explanation... I understand that we are finding the hypothenuse R via Pythagoras but it is especially the fact that the solution implies that $Rx = \frac{1}{2} \cdot F$. I cant seem to wrap my head around as to why though? And how do we know so?

Why is the force applied at end of rod by the axis of rotation $R$, at time, $t=0$, equal to $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$?

the question is regarding a problem in Rotational Motion for rigid bodies in AP Physics 1. I have tried to translate it as best as I could from Swedish. The problem is as follows:

"A homogeneous rod with length $L$ and mass $M$ is fixed to the horizontal axis. The rod can freely move around its axis. Initially, the rod is standing vertically. At t=0, a force F is applied at the end of the Rod (see image)

What is the magnitude of the Reaction Force, R, that the axis of rotation applies to the rod at t=0?"

An attempt to draw the Free-Body Diagram

I have tried to draw a Free-Body Diagram with the "Reaction Force" split into its x- and y-components. My teacher has published the solution, which is $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$ with no explanation... I understand that we are finding the hypothenuse R via Pythagoras but it is especially the fact that the solution implies that $Rx = \frac{1}{2} \cdot F$. I cant seem to wrap my head around as to why though? And how do we know so?

edited tags; edited title
Link
Qmechanic ♦
Qmechanic ♦
  • 212.9k
  • 48
  • 589
  • 2.3k

Why is the force applied at end of rod by the axis of rotation (R), at time, t = 0$t=0$, equal to $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$?

Source Link
Niko
Niko
  • 3
  • 1

Why is the force applied at end of rod by the axis of rotation (R), at time, t = 0, equal to $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$?

the question is regarding a problem in Rotational Motion for rigid bodies in AP Physics 1. I have tried to translate it as best as I could from Swedish. The problem is as follows: " A homogeneous rod with length L and mass M is fixed to the horizontal axis. The rod can freely move around its axis. Initially, the rod is standing vertically. At t=0, a force F is applied at the end of the Rod (see image)

What is the magnitude of the Reaction Force, R, that the axis of rotation applies to the rod at t=0? "

An attempt to draw the Free-Body Diagram

I have tried to draw a Free-Body Diagram with the "Reaction Force" split into its x- and y-components. My teacher has published the solution, which is $R=\sqrt{(( M\cdot g)^2+( \frac{1}{2} \cdot F)^2)}$ with no explanation... I understand that we are finding the hypothenuse R via Pythagoras but it is especially the fact that the solution implies that $Rx = \frac{1}{2} \cdot F$. I cant seem to wrap my head around as to why though? And how do we know so?