Greetings again, first I am not a student. I am just a silly man who likes to make sure I understand what I am talking about. Help me verify this:

Lets say I have a 1000 lb load. That would give me a mass equal to 1000/32 ft/s^2 or 31.25 (m=F/a)

Now lets say I want to accelerate this load with a move profile of 0 to 4 in/sec in 100msec. That would yield an acceleration of 40 in/sec^2.

To move this load at this acceleration vertically upward I would need a force of F-1000=31.25*40 or 2250 lbs.

I would like to connect this load to a rope wrap around a pulley of 2.5 inches in diameter. Radius = 1.25 Would the torque required at the shaft of the pulley be 2250*1.25 or 2812.5 inch-lbs?

If its this simple, understanding I am neglecting drags and frictions of some sort, why the need for moments of inertia?

Lastly, according the right hand rule of the cross product in figuring torque. Why is the torque(rotational force) pointing in the z direction. I mean the torque at my wheels of my car propel my car forward, but text books make it appear as though the torque is pointing in the direction of the axis of rotation wouldn't that tend to push the wheels outward away from the vehicle which it doesn't.

Thanks in advance.

 

plc_user1973,

Greetings again, first I am not a student. I am just a silly man who likes to make sure I understand what I am talking about. Help me verify this:

And I am a prickly old curmudgeon with too much time who usually knows what he is talking about.

Lets say I have a 1000 lb load. That would give me a mass equal to 1000/32 ft/s^2 or 31.25 (m=F/a)

OK, that is 1000lbf/32 = 31.25 slugs

Now lets say I want to accelerate this load with a move profile of 0 to 4 in/sec in 100msec. That would yield an acceleration of 40 in/sec^2.

Assuming constant acceleration, v = a*t ----> a = v/t ---> a = (1/3)/0.1 = 3.33 f/sec^2

To move this load at this acceleration vertically upward I would need a force of F-1000=31.25*40 or 2250 lbs.

The force needed is 31.25*(32+3.3) = 1104 lbf

I would like to connect this load to a rope wrap around a pulley of 2.5 inches in diameter. Radius = 1.25 Would the torque required at the shaft of the pulley be 2250*1.25 or 2812.5 inch-lbs?

No, 1104*1.25/12 = 115 lbf-ft

If its this simple, understanding I am neglecting drags and frictions of some sort, why the need for moments of inertia?

Does the pulley wheel have a significant moment of inertia?

Lastly, according the right hand rule of the cross product in figuring torque. Why is the torque(rotational force) pointing in the z direction. I mean the torque at my wheels of my car propel my car forward, but text books make it appear as though the torque is pointing in the direction of the axis of rotation wouldn't that tend to push the wheels outward away from the vehicle which it doesn't.

Torque is defined to be perpendicular to the force-radius plane. Torque not a linear force so it cannot force your wheels outward. The linear force at the end of the wheel propels your vehicle forward or reverse. That linear force by itself is not torque.

Ratch

 

I mean the torque at my wheels of my car propel my car forward

Not exactly. The torque at the axle is balanced by a force (reaction at the pavement) acting through a moment arm (radius of wheel). A force is linear, and in the direction of travel, which is why your car moves linearly. If that reaction force wasn't there, the torque would cause angular acceleration of the axle and wheel.

 

Ratch, my problem was right in the beginning. I should have used 384 in/sec^2 instead of 32 ft/sec^2 in my original post. I was in a hurry to rush to work and I wanted to leave this to you guys while I was away. Everything else I understand except, I am still not getting the the direction of the cross product of F and d in the calculation of torque. The books say its in the z direction(right hand rule). Torque is a rotational force as we know, but the books make it appear as though it a force that might be acting on an object in the z axis.

As for the pulley inertia as you asked. Lets say its 1lb pulley then I=(1/384)*(1.25)^2. In this system, I don't think it's significant in relation to the load.

P.S. I really am not a school student ie, high school, university, etc. I just enjoy learning what I can.

 

I suspect you're confusing the mental picture of an arrow representing a force vector with an arrow representing a torque vector. They're both vectors (well, one's a pseudovector), but they represent physically different things. A torque is not a force that you can plug into Newton's second law to deal with linear motion. There is a corresponding equation for angular motion, of course; torque is the analog of force, moment of inertia is the analog of mass, and angular acceleration is the analog of linear acceleration.

The easiest way to recognize this difference is that they have different physical units.

 

Not exactly. I can see the distinction between the force vector and the torque vector. The torque vector is perpendicular to the Force vector and moment area plane. The books make the torque vector point in the z axis. Why does it POINT in the z axis and not be an angled rotating arrow about the z axis?

Regards

 

plc_user1973,

Why does it POINT in the z axis and not be an angled rotating arrow about the z axis?

At present, torque is defined as the cross product of the radius and force, both of which are vectors. If the radius and force are both on the x-y plane, then of course the cross product will be on the z axis. What would be the vector equation of torque if it is defined your way?

Ratch

 

Get rid of the math for a moment and focus on what's going on physically. We know from experience (say, using a lever) that if you double the force applied around the fulcrum, you can lift double the load (for levers without friction). If we define the torque as the arithmetical product of distance and force, then you can see we have an (almost) essential measure of the ability to cause rotation about an axis. This works for forces at a right angle to the beam of the lever.

We also know from experience that as we change the angle between the force and the beam, the ability to cause rotation lessens. Thus, at 90° we get the "full" treatment, and as we "fall off" from 90°, the ability to lift something with the lever drops. It was determined experimentally that this relationship was (force)(distance)(sine of the angle).

That captures numerically what happens experimentally. No doubt all of us have used the principles at one time or another (here are my notes on "weighing" a trailer with a lever).

Since this torque is made by multiplying two vector quantities (position and force), it's possible that the result needs to be a vector (in general, it could be a scalar, vector, dyad/tensor, or something more complicated). What it "needs" to be is determined by utility and consistency. The most useful was to have it be a vector, as it needed to encode two things: the magnitude of the ability to cause rotation and the information about the direction of this rotation (since, if I ask you to calculate the result of applying a torque of 1 N*m to a rigid body, you can't give me an answer until I tell you about what axis the torque is applied and in which direction the body should rotate as a result of applying the torque).

For the direction, one could use a direction in the plane of the position and force vectors, but that has the failing of requiring the torque vector to continually change its direction in a dynamical case like a motor overcoming a constant friction in the bearings (when in fact our intuition would say that the required torque should be constant). The remaining "natural" choice was to make it parallel to the axis of rotation. The vector cross product was utilized for such things. I don't know enough about the history of vector algebra to say which was the dog and which was the tail though.

Hopefully, that makes things a bit clearer. It's mostly about being consistent and useful (and, though it may not appear that way, simple).

 

At present, torque is defined as the cross product of the radius and force

In my original post, I am pretty sure I alluded to this fact. I understand the math behind the cross product.

My issue is the math doesn't seem to convey the physical aspect of the rotational force.
I can already see you guys are smart so we don't have to talk it out like a text book. I am interested in intuitive understanding as a foundation.

I don't see the z direction as being intuitive other than it shows the axis of rotation, but to me it should be an arced curve around the axis of rotation.

Thanks for continuing to help out.

 

plc_user1973,

I don't see the z direction as being intuitive other than it shows the axis of rotation, but to me it should be an arced curve around the axis of rotation.

What is the vector equation for your proposed curve?

Ratch

 

Ratch,
do you really think after reading all my post's that I really think that is mathematically possible?

What is the vector equation for your proposed curve?

That type of equation is not possible to my understanding. Again, to me it is completely non-intuitive to have a rotational force point in a linear direction perpendicular to the plane of rotation.

I am not trying to irritate anyone here.

Thanks again

 

Ratch,
That type of equation is not possible to my understanding. Again, to me it is completely non-intuitive to have a rotational force point in a linear direction perpendicular to the plane of rotation.

I am not trying to irritate anyone here.

I don't think you're irritating anyone. If the explanations offered so far don't help, then you'll have to be satisfied with unintuitive nature of the direction of the torque vector or continue to study and understand. As I implied in a previous post, the definitions that survive (in a Darwinian sense) are useful and consistent; being intuitive is nice but not mandatory.

All of us are students and struggle with new ideas. I commend you for your attempts at understanding. It takes most folks until mid or late life to understand that what they really went to school for was to teach them how to teach themselves -- and that life is a continuous learning process.

 

plc_user1973,

Again, to me it is completely non-intuitive to have a rotational force point in a linear direction perpendicular to the plane of rotation.

What is the other type of direction other than linear at a point. What does does linear mean in this context.

When you have finished with torque, then check out angular velocity and angular momentum. They are other "moment" type quantities defined by the vector cross product.

Ratch

 

That type of equation is not possible to my understanding. Again, to me it is completely non-intuitive to have a rotational force point in a linear direction perpendicular to the plane of rotation.

Try to think of not as a rotational force (which it isn't), but as a moment. A moment acts around an axis, which you already understand because you would like to see it represented as a curved arrow around that axis. That "aroundiness" has a direction, which by convention we have defined by the right-hand rule. The direction is a lot easier to represent with a linear arrow along the axis than with a curved arrow around the axis. Maybe it's just physics graphical semantics, but it does the job. The more you play with your hand, wrapping it around the axis and looking at which direction your thumb is pointing, the more intuitive it becomes.
It it makes you feel any better, most of the time when I'm working on planar statics and dynamics problems as a practicing mechanical engineer, I'll represent moments by a curved arrow on the plane, not by a moment vector perpendicular to the plane. These are fairly simple problems, a far cry from the level of complexity found in robotics and other three-dimensional applications where the more rigorous approach is called for.