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Problem with resolution of forces

  • Thread starter sachin123
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  • #1
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I know of resolving forces into components along x and y axes.
But I have this issue.

Consider a bob suspended from a string.The string is not perpendicular to the horizontal,but inclined at some angle(due some force,say it is in a non inertial frame).
Let the angle be theta(with the vertical).
So which of these is correct:
a) m*g*cos(theta)=T or
b) m*g=T*cos(theta).
Why?
I have seen both equations used in books.

Thank You.
 

Answers and Replies

  • #2
Doc Al
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So which of these is correct:
a) m*g*cos(theta)=T or
b) m*g=T*cos(theta).
Why?
In this problem, I presume, there is only a horizontal acceleration.

The first equation (a) presumes that ΣF = 0 parallel to the string, which is a false. (There is a component of acceleration in that direction.) This is not correct.

The second equation (b) is derived from ΣF = 0 in the vertical direction. It is correct.

I have seen both equations used in books.
What book uses the (a) version?
 
  • #3
3,732
414
Each one can be correct. It depends on the equilibrium conditions in the problem. It's not a matter of one being "right" and the other "wrong".

a) represents the components along the string direction (radial direction for a circular motion or pendulum)
b) represents the components along the y (vertical direction)

So for a pendulum, (b) is not true as there is motion along the y axis as well as along the x axis but (a) is OK - there is no radial motion.
For the case of a "inertia force" acting horizontally and the bob being at rest, b) may be right.
 
  • #4
Doc Al
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Each one can be correct. It depends on the equilibrium conditions in the problem. It's not a matter of one being "right" and the other "wrong".
Good point!
So for a pendulum, (b) is not true as there is motion along the y axis as well as along the x axis but (a) is OK - there is no radial motion.
In general, neither (a) nor (b) is correct for a pendulum. (There is a radial component of acceleration.)
 
  • #5
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Thanks both of you.But I am not done yet.
I made up the problem for the sake of telling you what exactly my question was.
So,forgetting radial acceleration and inertial forces,or the problem itself,
what should we consider?

nasu:
Each one can be correct. It depends on the equilibrium conditions in the problem. It's not a matter of one being "right" and the other "wrong".
So how does it depend on the equilibrium conditions.Please explain.

I am just asking in general.If we just stick to resolution of forces into x and y components,will
we get the job done?
 
  • #6
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Oh I found a case when I can put this clearly.

A pendulum again,but suspended from a point free to revolve vertically).
At the angular position(with respect to vertical) ,
at which the reaction between the particle and the loop becomes zero,
m*g*cos(theta)=m(v^2)/r

Why isn't it m*g=(m(v^2)*cos(theta))/r?
 
  • #7
Doc Al
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Oh I found a case when I can put this clearly.

A pendulum again,but suspended from a point free to revolve vertically).
At the angular position(with respect to vertical) ,
at which the reaction between the particle and the loop becomes zero,
m*g*cos(theta)=m(v^2)/r

Why isn't it m*g=(m(v^2)*cos(theta))/r?
Can you give the complete statement of the problem? I'm unclear whether you are talking about a pendulum bob on a string or a mass on a circular loop.
 
  • #8
3,732
414
In general, neither (a) nor (b) is correct for a pendulum. (There is a radial component of acceleration.)
Correct, my mistake. Only at the extreme position (v=0) (a) may be true.
 
  • #9
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Can you give the complete statement of the problem? I'm unclear whether you are talking about a pendulum bob on a string or a mass on a circular loop.
A particle constrained to move along a smooth vertical loop of radius r.
 
  • #10
Doc Al
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A particle constrained to move along a smooth vertical loop of radius r.
OK. Nothing to do with a pendulum.

At the angular position(with respect to vertical) ,
at which the reaction between the particle and the loop becomes zero,
m*g*cos(theta)=m(v^2)/r

Why isn't it m*g=(m(v^2)*cos(theta))/r?
Consider the radial components of force (the only force acting is gravity) and apply Newton's 2nd law.
 
  • #11
3,732
414
nasu:
Each one can be correct. It depends on the equilibrium conditions in the problem. It's not a matter of one being "right" and the other "wrong".
So how does it depend on the equilibrium conditions.Please explain.

I am just asking in general.If we just stick to resolution of forces into x and y components,will
we get the job done?
There is no "in general". You can use any system of axes to resolve the forces.
If you use vertical-horizontal, you will have

Tcos(theta) and -mg

on the vertical axis.
If you choose radial-tangential axes, you will have

T and -mg*cos(theta)

on the radial axis.

You may also have various other components, from other forces.
Now when you write Newton's law for the specific axis, it reads:

Sum of components= m*a

So in both choices of axes, the two components that you are interested in will be equal only when a=0 for that axis.

mg=Tcos(theta) only if the vertical acceleration is zero.
mgcos(theta)=T only is the radial acceleration is zero.

In general, none of the accelerations MUST be zero so neither equality is true "in general".

Just write Newton's laws for the specific problem and you'll see what you get.
 
  • #12
Doc Al
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Why isn't it m*g=(m(v^2)*cos(theta))/r?
In case it wasn't clear, this (incorrect) equation is an attempt to look at vertical components of forces. The problem is that it doesn't properly take into consideration the tangential acceleration, which has a vertical component.

As nasu says, you can take any direction you like as long as you apply Newton's 2nd law correctly. But for many problems there is a direction that makes the problem easier to solve.
 
  • #13
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By the way,there is no tangential acceleration(zero).
So,this equation has to be right.
And,I have done what nasu has said.
I stuck to conventional x and y axes.
Haven't I?
Please help me out of this.
 
  • #14
Doc Al
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By the way,there is no tangential acceleration(zero).
What makes you think that? The mass is sliding down a frictionless surface, picking up speed as it goes.
 
  • #15
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Oh.Right.Thanks a lot Al and nasu.:)
 

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