Small block on a track - Finding Normal forces and

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Homework Help Overview

The discussion revolves around a physics problem involving a small block on a track, specifically focusing on finding normal forces and centripetal forces at a point on the track. Participants are exploring concepts related to circular motion and the forces acting on the block.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the block at point Q, including gravity and the normal force. There are inquiries about the nature of the forces required for circular motion and the specific term for the force directed towards the center of the circle. Some participants express uncertainty about the formula for centripetal force and how to apply it in this context.

Discussion Status

The discussion is active, with participants sharing their attempts to derive formulas and clarify concepts. Some guidance has been offered regarding the use of conservation of energy to find the speed at point Q, and there is an ongoing exploration of the relationship between centripetal force and normal force.

Contextual Notes

Participants mention difficulties with textbook resources and the need to clarify the speed at point Q for accurate calculations. There is a focus on ensuring that the correct values are used in the formulas being discussed.

nukeman
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Small block on a track - Finding Normal forces and...

Homework Statement



Here is the question. Can someone please help, since I missed the section of class that went over this, any my textbook does not cover this well. Look below at question 1, a) and b)

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Homework Equations





The Attempt at a Solution

 
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Which are the forces at point Q?
 


At point Q - Forces are Gravity down... Does it have tangential and radial?
 


The block is touching the track. Hence besides the weight there is also the ...
 


grzz said:
The block is touching the track. Hence besides the weight there is also the ...

Force of gravity and the normal force
 


Correct.
Now for a mass to perform circular motion, it needs a ...
 


grzz said:
Correct.
Now for a mass to perform circular motion, it needs a ...

Force?
 


What do we call this particular force?
 


I am not sure. Momentum force?
 
  • #10


Any mass performing circular motion needs a force towards the centre of the circle. This particular force is called ...
 
  • #11


Centripetal Force?
 
  • #12


CORRECT.

what is the formula for this centripetal force in terms of the speed at Q?
 
  • #13


grzz said:
CORRECT.

what is the formula for this centripetal force in terms of the speed at Q?

I am not sure on that... I can't find it in my book (awfull book). :(

?
 
  • #14


This forum is for helping and not for doing the work for you.
Try looking up 'centripeta force' on internet search engine. That is the way to learn.
 
  • #15


Did not mean to imply that.

F = MAc = MV^2/r

Is that it? what is the c ?
 
  • #16


It is ok. You are welcome to this forum!

You got the right formula.
A - acceleration, c- centripetal.

Now try to find the speed at point Q.
 
  • #17


This is where i have trouble.

When i plug everything in:

F = (.100)(2.00)^2 / (.150)

I get 2.66 ?
 
  • #18


Remember that we were talking about the mass when it is at point Q. So we are trying to find the centripetal force at point Q. But the speed you are using is 2m/s which is the speed at a higher point and not the speed at Q.

Try to find the speed at Q.
 
  • #19


I don't understand how I would change that formula to find the centripetal force at point Q
 
  • #20


The first part of the problem was to find the normal force at Q.

To do this we need to find the centripetal force at Q.

This centripetal force = mv^{2}/R.
Hence we need the speed v, at Q.
Try using the conservation of energy principle to find the speed at Q.
 
  • #21


Is the following correct?

A: (100)(2.00)^2/.150 = 2.66 N

B: For of gravity is added so, 2.66N - (.100) (9.8) = 1.68 N ?
 
  • #22


nukeman said:
Is the following correct?

A: (100)(2.00)^2/.150 = 2.66 N

B: For of gravity is added so, 2.66N - (.100) (9.8) = 1.68 N ?

Let us continue with the method:

Try to use the conservation of energy principle to find the velocity at Q.
 

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