# Effective Forces on a Rigid Body

1. Apr 9, 2004

### AngelofMusic

A question about Dynamics and drawing FBDs. In our book, we have an equation that says:

$$\Sigma M_O = \Sigma M_{O(effective)}$$

Most of the time, this basically translates into:

$$\Sigma M_O = \overline{I}\alpha$$. The left-hand side basically represents the sum of the moments of all the forces on the body and I usually have no trouble coming up with that. It's when the right-hand side includes moments caused by acceleration in addition to Iα where I get confused.

Diagram 1

In this case, $$\Sigma M_{O(effective)} = \overline{I}\alpha + m_Aa_Ar + m_Ba_Br$$, where the effective moments include the ones caused by the acceleration of the blocks attached to the pulley.

Is this only the case when the acceleration is due to objects directly connected to the disk?

For example, in the situation here: Diagram 2, although there is friction between the disk and the cart underneath (rolling without sliding), and the friction causes a centroidal acceleration in the disk in addition to its angular acceleration - the acceleration of the cart does not cause an effective moment in the disk.

The book doesn't quite explain the theory very well, so this is just my hypothesis. When would I have to take into account the acceleration of other objects when calculating the effective moment? Only when the rigid bodies are connected?

Any clarifications or additional explanations would be greatly appreciated!

2. Apr 10, 2004

### Staff: Mentor

Could you please elaborate on this concept of "effective moment"? I'm not familiar with the term and don't know what it's supposed to mean. (What class are you taking?)

The way I would apply Newton's 2nd Law for rotation is simple. You identify all the forces on a body, then calculate the torque they produce about the axis of interest. That total torque Τ = I α.

For your pulley with hanging masses example: The masses affect the pulley because the strings exert tension directly on the pulley. I would analyze each object in turn (pulley, mass1, mass2) first identifying all the forces acting on each object.

If you can give specific problems, perhaps I would understand better what your question is.

3. Apr 10, 2004

### AngelofMusic

According to the book, this is something called d'Alembert's Principle: The sum of the external forces acting on a rigid body are equivalent to the effective forces of the various particles forming the body. This book uses "moment" to represent torque, AFAIK.

The two questions listed were all sample problems from the book, but maybe you can explain it better.

For the 1st diagram, mass of the pulley is 50 kg and it has a radius of gyration of 0.4 m. It has a radius of 0.5 m. Block A is 20 kg and block B is 40 kg. Find the angular acceleration of the pulley.

In their explanation, they say: The effective forces reduce to the couple of I&alpha; and $$ma_A and ma_B$$.

For the 2nd diagram, a 0.24-m radius cylinder of mass 8 kg rests on a 3 kg cart. The system is at rest when a force of 10 N is applied. The cylinder rolls without sliding. What is the angular acceleration of the cylinder?

In this case, they say that the effective forces reduce to the couple I&alpha; and $$m\overline{a}$$ at the centre of the cylinder. (The cart just has a translational ma).

$$F_f(0.24) = \overline{I}\alpha, where F_f is the frictional force.$$
$$F_f = m\overline{a}$$

My question is - what makes these two situations inherently different? The fact that the pulley is directly connected to the masses in the first case? I didn't have to take into account the frictional force between the cylinder and the cart when calculating the effective moment/torque. (The stuff on the right-hand side of the equation).

4. Apr 11, 2004

### Staff: Mentor

Ah... Now I understand what you're talking about. I apologize for being so dense, but it has been decades since I've had to use Lagrangian mechanics and the principle of virtual work (one form of D'Alembert's principle). So... until I can dig up my old mechanics texts and review these ideas I won't say much lest I give you bad advice.

I hope that someone more up to speed on this will chime in.