Empty Box with Plumb Bob Sliding Down a Plane

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Homework Help Overview

The problem involves a plumb bob hanging inside an empty box that slides down a frictionless inclined plane at an angle of 30 degrees. The main question is to determine the angle the plumb bob makes with the vertical during the slide.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the forces acting on the plumb bob and the box, including tension and pseudo forces. There are attempts to draw free body diagrams and clarify the role of pseudo forces in a non-inertial frame.

Discussion Status

The discussion is ongoing, with various interpretations of the forces involved. Some participants are exploring the necessity of pseudo forces, while others suggest a simpler approach without them. Guidance has been offered regarding the components of forces and the need to find the angle of the plumb bob.

Contextual Notes

There is mention of the lack of prior knowledge about pseudo forces among some participants, and the discussion includes references to external resources for further understanding.

Karina
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Homework Statement


You hang a plumb bob inside an empty box. The empty box slides down a friction-less inclined plane. The inclined plane makes an angle of 30 degrees with the horizontal. What is the angle the plumb bob makes with the vertical during the slide?

Homework Equations


g pulled down on the string at 9.81 m/s^2... and the Tension = should be the same in the opposite direction

The Attempt at a Solution


I know that the box is accelerating down the slide and the string continues to exert an upward force on the plumb bob to balance the gravity acting on it. The plumb bob should hang an angle towards the back of the box. I tried to draw it out but I am not even sure if I have it right so far.
 

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Hi Karina. Welcome to PF.:)

Can you draw the free body diagrams of box and bob?
Do you know have knowledge about pseudo force?
 
Not sure if this will suffice? Is that correct? We haven't gone over that in class yet but I am aware that it's a fake force, like something we make up. Is that what I was thinking earlier in the explanation, a pseudo force?
 

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Karina said:
Not sure if this will suffice? Is that correct? We haven't gone over that in class yet but I am aware that it's a fake force, like something we make up. Is that what I was thinking earlier in the explanation, a pseudo force?

Yes pseudo force is a fake force.
pppppp.png

Figure 1 is the FBD of the box and Figure 2 is the FBD of the bob.
Here I am considering a reference frame which is moving with the box. As it is a non-inertial frame of reference so we have to consider a pseudo force. Can you find the magnitude of pseudo force acting on the bob?
 
Isn't the pseudo force = m2g+T?
 
Karina said:
Isn't the pseudo force = m2g+T?

No, why?
You have to find the magnitude of pseudo force in terms of ##m_{2}##,##g## and ##\theta##.
 
Pseudo force = m2gsintheta
 
Karina said:
Pseudo force = m2gsintheta
That's the magnitude, but you should also state the direction. Now combine it in an equation with the other forces on m2.
 
  • #10
Karina said:
Pseudo force = m2gsintheta

Yes that's correct.
I have also specified the direction of this pseudo force in #Post 4.

Kar.png


In this figure I have drawn the FBD of that bob. Consider force 'P' in the figure as pseudo force.
In figure I have also specified the the axis. Just find the components of forces along the X and Y axis and construct equation using second law.
For that you have to find the value of angle which I have mentioned. Can you find it?
 
  • #11
Don't intend to get into a fight about pseudo forces (I avoid them as much as I can).
But let me at least emphasize there's no need for them in this exercise:

The bob hangs from a rope, so it "feels" the tension of the rope, and the Earth is pulling on it straight down.
That's all the forces it "feels". If it's to accelerate down the slide at the same rate (##\ g\sin\theta\ ##) as the box (i.e. not swing around like puppet on a string), the net vector sum has to be ##mg\sin\theta\;##. In other words, the string has to compensate ( ## mg\;## straight down ) minus ( ##mg\sin\theta\;## downwards along the slope).

Much simpler...
 

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