Turning Effect of Forces

[JamSmith]

I was solving some online physics question papers.

 

Can anyone explain, what is Turning Effect of Force?

 

& also guide me how to solve this:

 

A uniform beam of 1 m is supported at the 50 cm mark. Given that a weight of 2 N hangs at the 30 cm mark, how far away from the pivot must another weight of 4 N be hung to balance the beam?

Edited March 27, 2017 by JamSmith

[swansont]

The concept here is torque, which is the ability to cause rotational acceleration, much like a linear force F causes linear acceleration.

 

T = rF sin(theta), where r is the distance from the pivot and F is the force. Theta is the angle between them; you maximize torque when the two are at right angles

[JamSmith]

The concept here is torque, which is the ability to cause rotational acceleration, much like a linear force F causes linear acceleration.

 

T = rF sin(theta), where r is the distance from the pivot and F is the force. Theta is the angle between them; you maximize torque when the two are at right angles

 

 

But if the beam isn't rotating (balanced), so I have to consider the clockwise moments equal the anti-clockwise moments?

[Sriman Dutta]

Moment of force is defined as the turning effect of a force on a rigid body about an axis of rotation.

 

Mathematically the moment of force or torque is a vector, and is equal to the vector cross product of force applied and the perpendicular distance.

[math]\tau = F X r [/math]

 

If the body under application of moments doesn't rotate, then according to the principle of moments, the body is in equilibrium due to the fact that the sum total of clockwise moments is equal to the sum total of anticlockwise moments.

Edited March 27, 2017 by Sriman Dutta

[swansont]

 

But if the beam isn't rotating (balanced), so I have to consider the clockwise moments equal the anti-clockwise moments?

 

 

Yes, exactly.

[Velocity_Boy]

I was solving some online physics question papers.

 

Can anyone explain, what is Turning Effect of Force?

 

& also guide me how to solve this:

 

A uniform beam of 1 m is supported at the 50 cm mark. Given that a weight of 2 N hangs at the 30 cm mark, how far away from the pivot must another weight of 4 N be hung to balance the beam?

 

 

The equation is referring to the process of "torque" which is the amount of twisting force (t) applied to an object along its longitudinal axis. It is just a different direction of force (f). NO different than how force is used on mass (m) to cause acceleration.

[swansont]

 

 

The equation is referring to the process of "torque" which is the amount of twisting force (t) applied to an object along its longitudinal axis. It is just a different direction of force (f). NO different than how force is used on mass (m) to cause acceleration.

 

 

Yes, it's different, because it depends on where the force is applied. For linear motion that distinction doesn't come up — you usually assume it has no effect.

[JamSmith]

Moment of force is defined as the turning effect of a force on a rigid body about an axis of rotation.

 

Mathematically the moment of force or torque is a vector, and is equal to the vector cross product of force applied and the perpendicular distance.

[math]\tau = F X r [/math]

 

If the body under application of moments doesn't rotate, then according to the principle of moments, the body is in equilibrium due to the fact that the sum total of clockwise moments is equal to the sum total of anticlockwise moments.

I got your point, I can also follow this right?

 

Moment (Nm) = Linear force applied at a point (N) X Perpendicular distance between applied point and pivot point (m)

 

We can then specify the rotational equivalent of Newton's laws:

1. If the moments on a rotating object are balanced then the object remains stationary or at constant angular velocity.

2. If the moments on a rotating object are unbalanced then the object will have an angular acceleration.

3. Moments are balanced by equal and opposite moments

 

The rotational equivalent of Newton's first law is often called "The principle of moments", which is "if the sum of the clockwise moments is equal to the sum of the anticlockwise moments, the object does not rotate or rotates at constant angular speed".

[Sriman Dutta]

You are correct. Note that the torque is a vector and has a direction perpendicular to that of force and radius vectors.

Translational motion

Newton's law: The force exerted by a body is directly proportional to the rate of change of linear momentum.

Equation: [math]F=ma[/math]

Equations of motion:

1. [math]v=u+at[/math]

2. [math]S=ut+\frac{1}{2}at^2[/math]

3. [math]v^2=u^2+2aS[/math]

 

Rotational motion

Newton's law: The torque on a body is directly proportional to the rate of change of angular momentum.

Equation: [math]\tau = I\alpha[/math]

Equations of motion:

1. [math]\omega_f = \omega_i + \alpha t[/math]

2.[math]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/math]

3. [math]\omega_f^2 = \omega_i^2 +2\alpha \theta [/math]

[Bender]

3. Moments are balanced by equal and opposite moments

I don't like how you formulated this one, as it can be confusing and could confirm common misconceptions about the third law.

More accurately: when object A exerts a moment on object B, object B simultaneously exerts an equal but opposite moment on object A.

[JamSmith]

You are correct. Note that the torque is a vector and has a direction perpendicular to that of force and radius vectors.

Translational motion

Newton's law: The force exerted by a body is directly proportional to the rate of change of linear momentum.

Equation: [math]F=ma[/math]

Equations of motion:

1. [math]v=u+at[/math]

2. [math]S=ut+\frac{1}{2}at^2[/math]

3. [math]v^2=u^2+2aS[/math]

 

Rotational motion

Newton's law: The torque on a body is directly proportional to the rate of change of angular momentum.

Equation: [math]\tau = I\alpha[/math]

Equations of motion:

1. [math]\omega_f = \omega_i + \alpha t[/math]

2.[math]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/math]

3. [math]\omega_f^2 = \omega_i^2 +2\alpha \theta [/math]

I was wondering about this solution. Well, thank you Sriman Datt.

I don't like how you formulated this one, as it can be confusing and could confirm common misconceptions about the third law.

More accurately: when object A exerts a moment on object B, object B simultaneously exerts an equal but opposite moment on object A.

Oh ok, I am grateful to you for your more precise discussion.

[studiot]

 

 

Bender

I don't like how you formulated this one, as it can be confusing and could confirm common misconceptions about the third law.

More accurately: when object A exerts a moment on object B, object B simultaneously exerts an equal but opposite moment on object A.

 

Jamsmith

Oh ok, I am grateful to you for your more precise discussion.

 

Does it work like this?

 

How does it apply to a catherine wheel or a lawn sprinkler?

[Bender]

as written

[studiot]

 

as written

 

Go on?

Edited March 30, 2017 by studiot

[Bender]

If the nozzle exert any moment on the water drop, that water drop will exert an equal and opposite moment on the nozzle. While this moment depends on the geometry of the nozzle, it is most likely negligible for the dynamics of the sprinkler.

[studiot]

If the nozzle exert any moment on the water drop, that water drop will exert an equal and opposite moment on the nozzle. While this moment depends on the geometry of the nozzle, it is most likely negligible for the dynamics of the sprinkler.

 

 

In the water sprinkler, the water exits tangentially to the spin circle of the spinner.

Consequently object A (the water) exerts a moment on object B (the spinner) , about its central mounting.

 

However, as you have just observed, the spinner exerts no moment on the exiting water.

 

Normally the spinner incorporates balanced pairs of nozzles so actually generating a couple rather than a simple moment.

 

In the catherine wheel, there may be only one exhaust and therefore a simple moment, rather than a couple, is generated.

[Bender]

Except that the water does not exert a moment on the spinner, but a force. This force results in a moment combined with a distance from the centre of rotation. An easy way to see this is when you draw a moment diagram of the spinner. You will notice how the moment in the crosssection of the spinner goes to zero at the nozzle. How can the internal moment be zero if the water supposedly exerts a moment at that point?

 

If the water actually exerted a moment on the spinner, it would not matter at what distance from the centre, or in what direction, the water was going. A moment is independent of its point of exertion.

[studiot]

Except that the water does not exert a moment on the spinner, but a force. This force results in a moment combined with a distance from the centre of rotation. An easy way to see this is when you draw a moment diagram of the spinner. You will notice how the moment in the crosssection of the spinner goes to zero at the nozzle. How can the internal moment be zero if the water supposedly exerts a moment at that point?

 

If the water actually exerted a moment on the spinner, it would not matter at what distance from the centre, or in what direction, the water was going. A moment is independent of its point of exertion.

 

And I thought perhaps you were using the European convention, not the American one, which is confusing.

 

 

A couple has the property stated in your last line.

 

 

The moment of a single force is the different about nearly every point in a plane containing the line of action of the force.

It is zero at every point along the line of action.

 

A couple is the same about every point in the plane in which it acts.

 

There does not have to be any material body at the point of action of a moment or couple.

[Bender]

Conventions can be messy, which is why I prefer not to use distinctions between moment/couple/torque. I don't even know which convention is European and which American. As an engineer, I mostly use the Dutch translation of "torque", which is "koppel", which literally translates to "couple", all of which are used for pure "moments", but in some books only refer to pure moments that are caused by exactly two equal and opposite forces.

To be honest, I have only encountered problems with these conventions discussing with physicists, who tend to be more rigorous in the use of terminology .

 

And that doesn't even touch the confusion with "moment of inertia", which in Dutch is often simply "moment", not to be confused with "impulsmoment" which is Dutch for angular momentum...

 

EDIT: on second thought: under what definition would you say the water exerts a "moment" on the sprinkler? In that case, I would always specify the involvement of a force to avoid confusion.

Edited March 30, 2017 by Bender

[studiot]

Jamsmith wisely asked about "the turning effect of forces" and it is quite true that there are several different and distinct mechanical effects associated with turning.

 

So it would seem to me highly sensible and desirable to identify each of these effects with a unique name.

 

Unfortunately too many mix up the available different terms to create the general confusion, particularly for beginners, that holds today.

 

So Perhaps a little history might help?

 

Ca 250 BC The mechanics of turning effects was known to the ancient world for example the principle of levers attributed to Archimedes.

 

1725 The term moment was introduced and formally defined by Varignon in his book 'Nouvelle Mechanique.' “The moment of a force, P, about a point O is defined as the product of that force into the perpendicular OM drawn to its line of action from O, this perpendicular being reckoned positive or negative according as it lies to the left or right of the of the direction of P." Varignon's theorem holds to this day and may be found on Wikipedia.

 

1750 – 1804 St Vennant investigated the torsion of prismatic bars and posed St Vennant’s Problem. He did not however introduce new concepts in turning.

 

1804 -1806 Poinsot published his book 'Elements de Statique' and the theorem that bears his name. This introduced two things. He defined and introduced the term ‘couple’ and the theorem which states that in 3 dimensions any system of forces may be reduced to a single force plus a couple, in a plane perpendicular to the line of action of the force. He clearly defined his couple to exist in a plane.

 

1912 Lamb, one of the most prominent applied mathematicians of his time, proposed that the term ‘torque’ be introduced to replace ‘couple’ Lamb 'Statics' p52. “Since a couple in a given plane is for the purposes of pure statics sufficiently defined by its moment, it has been proposed to introduce a name torque or twisting effect which shall be free from the irrelevant suggestion of two particular forces.” This suggestion was not, however generally adopted. Indeed the three most influential texts ( in this subject) of that era and since carried on as before.

 

1926 Love ‘A Treatise on the Mathematical Theory of Elasticity’

 

1936 Southwell ‘Theory of Elasticity’ These both refer to ‘Torsional Couples’ for the 3D effects described in St Vennant’s Problem.

 

1934 Timoshenko published the third standard text, ‘Theory of Elasticity’ and clearly establish torque in this 3D role. In fact most authors in the second part of the 20th century have followed the notation set by Timoshenko in elasticity.

 

I haven't ventured beyond the first half of the 20th century because nothing new has been added since.

 

It does bring out one other source of confusion. The difference between twist and turn, which is even less often correctly stated.

I usually try to associate the Ts Torque, Torsion and Twist.

Edited March 30, 2017 by studiot

[Bender]

Thanks for the interesting historical overview.

 

In my experience, students are least confused when I simply avoid any convention and explain all the different terms are mixed. When I do not refer to a force, it is a "pure moment" (sometimes stressing the "pure" if the situation asks for it); when the moment is caused by a force at a distance, I specify.

[studiot]

I think a big problem for students is relating theory to practice.

 

A few practical examples can bring out the important principles and provide a lot of fun.

 

A hammer, nails, a wooden post and a lamina to nail to the post can demonstrate the difference between what happens when you apply a moment, (via a single force) and when you apply a couple.

 

A more sophisticate experiment is as follows.

 

Consider sharpening a chisel of thickness 4mm.

 

The chisel is applied tangentially to a rotating grindstone of diameter 200mm and coefficient of friction 0.6.

The chisel is pressed against the surface of the stone with a force of 10N and the frictional force is resisted by axial support at the other end of the chisel.

 

Now the normal force pressing against the grindstone produces a normal 10N reaction on the chisel blade.

Neither produce moments in the stone or chisel.

 

The frictional force of the blade against the stone is 6N and acts tangentially so it applies a moment of 6*0.2 = 1.2 N-m, opposing the rotation of the stone.

 

The counter frictional force acting on the chisel blade, also 6N, is balanced tangentially by the axial support force supplied to the other end of the chisel.

But these two forces are not coaxial.

 

The support acts along the centreline 2mm from the contact surface.

The friction acts in the opposite direction along the contact surface.

These two forces form a couple about the tending equal to 6*.002 = .012 N-m

 

So the turning effects are not equal by a long margin.

Edited March 30, 2017 by studiot

[JamSmith]

Jamsmith wisely asked about "the turning effect of forces" and it is quite true that there are several different and distinct mechanical effects associated with turning.

..........

Thanks for detailed information and clarification.