Article 34612 of rec.arts.disney: Newsgroups: rec.arts.disney,sci.physics Path: boulder!tali.hsc.colorado.edu!csn!magnus.acs.ohio-state.edu!math.ohio-state.edu!cs.utexas.edu!uunet!mnemosyne.cs.du.edu!nyx10!rito From: rito@nyx10.cs.du.edu (robert ito) Subject: The Physics of Disney's Beauty and the Beast Message-ID: <1994Apr1.001856.17546@mnemosyne.cs.du.edu> Followup-To: rec.arts.disney Summary: Could Belle catch the Beast? X-Disclaimer: Nyx is a public access Unix system run by the University of Denver for the Denver community. The University has neither control over nor responsibility for the opinions of users. Sender: usenet@mnemosyne.cs.du.edu (netnews admin account) Organization: Nyx, Public Access Unix at U. of Denver Math/CS dept. Date: Fri, 1 Apr 94 00:18:56 GMT Lines: 332 Xref: boulder rec.arts.disney:34612 sci.physics:61866 One of the more dramatic moments in Disney's "Beauty and the Beast" occurs when Gaston stabs the Beast, and the Beast starts to fall off the castle wall to his impending death. Belle quickly reaches for the Beast, grabs him, and breaks his fall. But you have to wonder if this slender (if tall) young woman could really catch and hold someone as large as the Beast. Here we'll try to analyze the physics of this situation and answer that question. But first, my thanks to David Uy, Amberle 'Meg' :-) Ferrian, Jeff Wilson and Tim Pickett for catching and correcting all the mistakes I'd originally made because of my rusty physics, and for all their very helpful comments. This isn't exactly a nice, neat textbook problem, so first let's write down our: =============== ASSUMPTIONS =============== 1. The Beast's height = 7'0" = 213.4 cm. It's not too easy to estimate his height, since his usual posture is hunched over, but if you think back to when Belle and the Beast are dancing during the ballroom scene, that should give you the best idea of their relative heights. (For reference, use Amberle's estimate that Belle is about 5'7".) 2. Beast's weight = 600 lb = 272.2 kg. The Beast, especially in his upper body, is very broad and very massive. Now, Shaquille O'Neal is about the same height, and he weighs about 300 pounds, so based on the Beast's proportions my guess is that he weighs at least twice what Shaq does. Also, Jeff suggested approximating the Beast as a right circular cone. I tried something a little different: Treat the Beast as two cones, base to base, with the bases meeting at his center of mass (see #3). Let h = height of the shorter cone = height(Beast) / 3. Then the combined volume of the two cones is pi * r^2 * h. Assume that r = 35 cm = 13.78", and that the Beast is as dense as water. Then v = 2.736 x 10^5 cm^3, and mass = 273.6 kg, which is almost the same as what I'd guessed above. So 600 pounds is in the right ballpark. 3. The Beast's center of mass (c.m.) is at his chest level, or 2/3rds of his height, or 4'8", or 142.2 cm. The Beast is, after all, definitely top-heavy, so his c.m. is above his mid-point. Also, when analyzing forces and torques later, we'll treat the Beast as a point mass located at his c.m. 4. Belle grabbed the Beast at his chest level, or at his center of mass, as Amberle verified from her laserdisc. Also, Belle's force is exerted entirely in the +x direction, so there's no component of force in the +y or -y direction. 5. This is a static equilibrium problem. That is, the Beast is stationary (at least momentarily) when Belle catches him, so he's not accelerating in any direction or around any axis. This means that the vector sum of all external forces on the Beast is zero, and the same is true for the external torques. 6. The Beast can be treated as a rigid beam. 7. The Beast is hinged to the wall. It looks like the Beast has his feet on the edge of the balcony, with his toes curled up and over the edge. That should keep his feet from slipping down the outer vertical surface of the balcony, or across the flat horizontal (and slick-with-rain) surface. Ordinarily we'd have to worry about the frictional forces that would keep his feet from slipping, but that shouldn't be necessary here. 8. The Beast is leaning at about 30 degrees from the vertical. (This was measured by Amberle off her laserdisc.) =============== CONVENTIONS =============== Just to keep things straight: 1. The +y direction is up. The +x direction is directed horizon- tally and perpendicularly inwards towards the balcony (and will be drawn as directed towards the right, to follow convention). And +z is towards us. 2. A torque acting in the clockwise direction is negative, and counterclockwise is positive. (This agrees with the vector equation for torques, T = r x F.) 3. Angles are in degrees. ================ CALCULATIONS ================ 1. THE FORCES ON THE BEAST For the forces on the Beast we have these three vectors: * W(beast), the force of gravity on the Beast. Origin = c.m.(beast) (the Beast's center of mass) Direction = straight down, or in the -y direction Magnitude = 272.2 kg * 9.807 m/s^2 = 2669.5 newtons (N) (1 newton = 1 kg-m/s^2) * F(belle), the force Belle exerts on the Beast. Origin = c.m.(beast) Direction = +x direction Magnitude = TBD (this is what we're trying to solve for) * F(wall), the force exerted on the Beast by the wall/balcony. Origin = point of contact between the wall/balcony and the Beast's feet Direction = in the -x / +y direction, at some TBD angle Magnitude = TBD If we draw a diagram of the force vectors acting on the Beast, it looks like this: _ \ \ F(wall) \ (c.m.)-----> F(belle) | | | W(beast) v Since this is a static situation, the vector sum of these forces is zero, and: F(belle) + W(beast) + F(wall) = 0 2. ANALYSIS OF THE COMPONENTS OF THE FORCES For the force components in the y-direction, the sum is also zero. If we write the y-component of F(wall) as F(wall,y), then: 0 = W(beast) + F(wall,y) = ( 272.2 kg * - 9.807 m/s^2 ) + F(wall,y) or: F(wall,y) = 272.2 kg * 9.807 m/s^2 = 2669.5 N For the force components in the x-direction: F(belle) + F(wall,x) = 0 But as David made clear, we really can't say much about either of these yet. So we have one equation with two unknowns, and we can't calculate F(belle) this way. 3. ANALYSIS OF THE TORQUES Since we can't calculate F(belle) by analyzing the forces, let's look at the torques. For the magnitudes of the torques involved, David pointed out that we can write: T = the net torque around the origin = 0 = sum (over i) [ r(i) * F(i) * sin(theta(i)) ] = ( 1.422m * W(beast) * sin(30) ) + ( - 1.422m * F(belle) * sin(60) ) + ( 0 * F(wall) * sin(??) ) In this case, the origin, or the point through which the axis of rotation runs, is where the Beast has his feet on the wall/balcony. And note that since we're working with magnitudes here, and not vectors, convention #2 makes the terms add up correctly (to zero). To try to clarify the equations above: Suppose we apply a force F(i) to a particle m(i) at a distance r(i) away from the origin O (as shown in the diagram here). _ ^ / : / <-- F(i) F(i) : / (perp.)--> : / : / theta(i) :/ m(i)<-------------------------(O) r(i) What's the magnitude of the torque on m(i) due to F(i)? Among other things, it depends on the angle theta(i) between r(i) and F(i). If F(i) is parallel to r(i), there's no torque, as such a force wouldn't cause m(i) to rotate around O. So, for torques we want the component of F(i) that's perpendicular to r(i). That's F(i) (perp.) above, and that has a magnitude of F(i) * sin(theta(i)). If F(i) and r(i) are parallel, then theta(i) is 0 or 180, and sin(theta(i)) = 0, and there's no torque. If theta(i) = 90, F(i) is perpendicular to r(i), and all of the force contributes to the torque. Returning to the torque equation, let's first cancel out the r(i)'s, since W(beast) and F(belle) are both acting on the Beast's center of mass, and r(i) is the same for both contributions to the torque. Also, r(i) for F(wall) is 0, so F(wall) doesn't contribute to the torque. Rearranging gives us: W(beast) * sin(30) = F(belle) * sin(60) or: F(belle) = W(beast) * sin(30) / sin(60) = 2669.5 N * 0.5 / 0.866 = 1541.2 N (or 157.2 kg force!) Now, at the very VERY most, Belle weighs 60 kg, or 132.3 pounds (while she's tall, she's also rather slender), so Belle's catching of the Beast is truly a Herculean feat. After all, she's pulling over 2 1/2 times her weight, if not three times her weight. And as Amberle pointed out, this is only the force needed to keep the Beast from falling. Belle would have to exert an even greater force to stop the Beast's fall, or to pull the Beast back up to the balcony. The calculation of that force, and the work done by Belle, is left as an exercise to the reader :-). Also, if you look again at the equations above, you'll see that if Belle grabs the Beast below his center of mass, F(belle) will have to be larger - and vice versa. And if Belle had applied her force perpendicular to the Beast's axis (instead of at the smaller angle that was assumed), she would've made things easier for herself. In that case, F(belle) = 1334.8 N or 136.1 kg force. 4. A WORST CASE SCENARIO Jeff noted that we can rewrite the above as: F(belle) = W(beast) * sin(theta) / sin(90 - theta) = W(beast) * sin(theta) / cos(theta) = W(beast) * tan(theta) where theta is the angle the Beast makes from the vertical. As theta increases, so does F(belle). For a worst case scenario, let's say that the Beast is 7'6" and 300 kg (as Amberle had initially assumed), and that he's leaning out at 45 degrees. Then: F(belle) = W(beast) * tan(45) = 300 kg * 9.807 m/s^2 * 1 = 2942.1 N (or 300 kg force!) which is a lot worse than the previous number we came up with. This is clearly impossible. Also, let's guess that at most Belle could exert an F(belle) equal to her weight, which we'll take to be 60 kg. Then, using W(beast) = 272.2 kg again, we can calculate the largest angle at which Belle could support the Beast, which is: = arctan( F(belle) / W(Beast) ) = arctan( 60 / 272.2 ) = 12.4 degrees ===================== FURTHER EXERCISES ===================== And, just for fun :-), here's some further exercises for the reader. 1. Remember the scene where the Beast grabs Gaston by the throat and then holds him out at arm's length in mid-air? What's the torque due to Gaston's weight? It must be large, although this is probably not as impressive a feat as Belle's catching of the Beast. (Assume that Gaston weighs 250 pounds, or 113.4 kg, and that the Beast's arm is about a meter long ...) 2. What force did Belle exert when she swung that tree branch and knocked the wolf off Philippe's back? An average wolf is pretty heavy - about 75 pounds or 34.0 kg - so this is no mean feat, obviously. One way to think about this, I'd guess, is to consider two problems that are similar (in one, you swing a tree branch, in the other a baseball bat): (1) the change in momentum imparted to the wolf by Belle, which isn't too hard to calculate (with an assumption or two), and (2) the delta(m*v) needed to knock a major league fastball over the fence in dead center field. Bob "Yojimbo" Ito Chief Editor, Disney Comics in The Future Disney Cabinet, bodyguard to the FDC walkaround Belle, and Jasmine, and part-time Walkaround Grumpy. "Is it, err, Mildred? Okay, no. How 'bout - Diana? Rachel?"