Small block on a track - Finding Normal forces and

  • Thread starter nukeman
  • Start date
  • #1
655
0
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)

v5fhuv.jpg




Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
998
15


Which are the forces at point Q?
 
  • #3
655
0


At point Q - Forces are Gravity down... Does it have tangential and radial?
 
  • #4
998
15


The block is touching the track. Hence besides the weight there is also the ...
 
  • #5
655
0


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

Force of gravity and the normal force
 
  • #6
998
15


Correct.
Now for a mass to perform circular motion, it needs a ...
 
  • #7
655
0


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

Force?
 
  • #8
998
15


What do we call this particular force?
 
  • #9
655
0


I am not sure. Momentum force?
 
  • #10
998
15


Any mass performing circular motion needs a force towards the centre of the circle. This particular force is called ...
 
  • #11
655
0


Centripetal Force?
 
  • #12
998
15


CORRECT.

what is the formula for this centripetal force in terms of the speed at Q?
 
  • #13
655
0


CORRECT.

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

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

?
 
  • #14
998
15


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
655
0


Did not mean to imply that.

F = MAc = MV^2/r

Is that it? what is the c ?
 
  • #16
998
15


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
655
0


This is where i have trouble.

When i plug everything in:

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

I get 2.66 ?
 
  • #18
998
15


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
655
0


I dont understand how I would change that formula to find the centripetal force at point Q
 
  • #20
998
15


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[itex]^{2}[/itex]/R.
Hence we need the speed v, at Q.
Try using the conservation of energy principle to find the speed at Q.
 
  • #21
655
0


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
998
15


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.
 

Related Threads on Small block on a track - Finding Normal forces and

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
4K
Replies
5
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
10
Views
13K
Replies
2
Views
4K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
2K
Replies
9
Views
1K
Top