Problem with resolution of forces

In summary, the conversation discusses the use of equations for resolving forces into components along x and y axes, particularly in cases involving pendulums or objects suspended from a string. The speakers mention the importance of considering equilibrium conditions and the use of different axes to resolve forces. They also discuss how the equations for resolving forces may vary depending on the specific problem and its components. Overall, the conversation highlights the importance of correctly applying Newton's 2nd law and considering all components of forces in a problem.
  • #1
sachin123
121
0
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.
 
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  • #2
sachin123 said:
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
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
nasu said:
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
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
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
sachin123 said:
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
Doc Al said:
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
Doc Al said:
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
sachin123 said:
A particle constrained to move along a smooth vertical loop of radius r.
OK. Nothing to do with a pendulum.

sachin123 said:
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
sachin123 said:
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
sachin123 said:
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
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
sachin123 said:
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
Oh.Right.Thanks a lot Al and nasu.:)
 

FAQ: Problem with resolution of forces

What is the problem with resolution of forces?

The problem with resolution of forces is that it is not always possible to accurately determine the individual forces acting on an object when multiple forces are acting on it. This is because forces can act in different directions and cancel each other out, making it difficult to determine the exact magnitude and direction of each force.

How is the problem with resolution of forces typically solved?

The problem with resolution of forces is often solved using vector analysis, which involves breaking down all the forces acting on an object into their individual components and then using mathematical equations to determine the resultant force and its direction. This allows for a more accurate determination of the forces acting on an object.

What factors can affect the accuracy of resolving forces?

There are several factors that can affect the accuracy of resolving forces, such as the precision of the measurements taken, the presence of external forces not accounted for, and the complexity of the object and its motion. In addition, human error and approximations made during calculations can also impact the accuracy of the results.

What are some real-world applications of resolving forces?

Resolving forces is an important concept in various fields such as physics, engineering, and biomechanics. It is used to analyze the forces acting on structures, machines, and human bodies, and to make predictions about their motion and stability. It is also essential for designing and optimizing structures and systems for maximum efficiency and safety.

What are some common misconceptions about resolving forces?

One common misconception about resolving forces is that all forces must act in the same direction in order to be resolved. In reality, forces can act in any direction and still be resolved using vector analysis. Another misconception is that the resultant force is simply the sum of all the individual forces, when in fact it takes into account their magnitudes and directions. Finally, some may mistakenly believe that resolving forces is only necessary for stationary objects, when in fact it is equally important for objects in motion.

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