# Planar Kinetics of a Rigid Body

1. Oct 11, 2004

### jjiimmyy101

QUESTION: The 50 kg uniform crate rests on the platform for which the coefficient of static friction is us = 0.5. If the supporting links have an angular velocity w = 1 rad/s, determine the greatest angular acceleration they can have so that the crate does not slip or tip at the instant theta = 30 as shown in the attachment.

I have no clue where to even begin. :yuck:

I need a hint or hints to start the problem. Please...anyone.

#### Attached Files:

• ###### pic 17-49.JPG
File size:
6.6 KB
Views:
194
2. Oct 11, 2004

### Soveraign

Ew. Well, start thinking about converting your angular acceleration to what that means in the x and y directions when the angle is 30 degrees. Keep in mind with regard to friction that 1) the y-acceleration will affect the normal force of the crate, thus affecting the frictional force and 2) the x-acceleration can be used to figure what lateral force the beam is creating on the crate, and since they mentioned it, 3) how the existing angular velocity affects the x and y accelerations on that beam due to the circular motion. When you have all the contributions to x and y acceleration, they can simply be added in the x and y directions independently so you should be able to get some good relations from the individual parts.

So, with regard to the "tipping over" part, similarly to above, you need to consider how the x and y accelerations contribute to the stability of this crate. I suppose were I to do this problem, I would start out by treating the two situations separately and then looking at which angular acceleration is smaller for the final answer.

Man, this sounds like an evil engineering dynamics problem.

3. Oct 11, 2004

### jjiimmyy101

ax = at = (angular acceleration) * radius

ay = an = (angular velocity) * radius

Am I getting there? How's my FBD of the crate look?

#### Attached Files:

• ###### fbd.JPG
File size:
7.2 KB
Views:
184
4. Oct 12, 2004

### Soveraign

The accelerations due to 1) the angular velocity and 2) the angular acceleration can each be broken down into x and y components.

Looking at the angular velocity case, you know that there is a centripetal acceleration in the direction parallel to the supporting links. This follows the familiar $a = v^2/r$ where $v = \omega r$. Thus, $a_x = a \cos(\theta)$ and $a_y = a\sin(\theta)$ where $\theta$ is the angle measured from the horizontal (i.e. 60 degrees).

So, putting these together you get the x and y contributions to acceleration due to the angular velocity as:

$a_x = \omega^2 r \cos(\theta)$ and $a_y = \omega^2 r \sin(\theta)$.

Now, the contributions from angular acceleration would be due to a traditional acceleration vector in the direction perpendicular to the supporting links with magnitude $a = \alpha r$. You can break down the x and y components after thinking about what direction this vector points.

At this point I'd like to stress that I may not be solving this in a way that your instructor intends. The overall idea is to find the total of the x and y accelerations on the box and then see how this affects the forces. For example, the normal force N would be the mass of the box times the sum of the accelerations due to gravity, angular velocity, and angular acceleration (keeping in mind some of these are negative). Also, it has been a while since I've had mechanics so I might be missing an easier way to do this.

5. Oct 12, 2004

### jjiimmyy101

angular acceleration:

$ax = \alpha r cos30$
$ay = \alpha r sin30$

Then:

$ax(total) = \sqrt(2^2 + (3.4641 \alpha)^2)$
$ay(total) = \sqrt(3.4641^2 + (-2 \alpha)^2)$

So then I should solve these equations:

$$\sum Fx = max$$ = -0.5*N = 50 *($\sqrt(4 + 12 \alpha^2)$
$$\sum Fy = may$$ = -W + N = 50 *($\sqrt(12 + 4 \alpha^2)$

Am I doing it right? I don't want to go any further if I'm wrong because those look like two tough equations to solve (very tedious). Thanks.

6. Oct 13, 2004

### Soveraign

Looks like you have the right idea for angular acceleration, although the vector is 30 degrees below the horizontal, so that would be a negative 30 degrees.

This would make the total accelerations:

$a_{xf} = \alpha r \cos(-30) + \omega^2 r \cos(30)$
$a_{yf} = g + \alpha r \sin(-30) + \omega^2 r \sin(30)$

(where g = -9.8, note y acceleration due to angular velocity is opposite direction to the other two and I have chosen down to be negative)

and thus, for example, the normal force $N = m a_{yf}$ leads to a frictional force of
$F_{fric} = \mu m a_{yf} = \mu m (g + \alpha r \sin(-30) + \omega^2 r \sin(30))$

The lateral force applied to this mass then becomes
$F_{lat} = m a_{xf} = m (\alpha r \cos(-30) + \omega^2 r \cos(30))$

so to figure out the maximum angular acceleration you can apply before the crate slides, set $F_{fric} = F_{lat}$ and solve for $\alpha$ (I think :uhh: ).

7. Oct 16, 2004

### jjiimmyy101

Thanks and sorry for not responding sooner.

$\alpha = -1.651$ is the answer I get when I solve $F_{fric} = F_{lat}$
but the answer is supposedly $\alpha = 0.587$

I was just wondering if you're taking into consideration if it tips at this instant as well as slips. I don't know what's going on and I sure wouldn't have gotten this far without you, so I'm just throwing out suggestions now. I'm at the brink of just giving up on this question...it's starting to annoy me.