Word Problem Wednesday: A Weighty Question!

Are you smarter than a 3^rd grader? Our curriculum writer has posed a weighty question!

One robot can lift and carry 900 pounds. A second robot can lift and carry 1,800 pounds. If the robots work together, can they lift a boulder that weighs 2,500 pounds?

Assume the robots are sci-fi mechanical humans and not specialized lifting machines like cranes or drones. 

Answer waaaay down below. BTW, as we explain our answers, actual experience lifting heavy packages, like furniture, will help you visualize the explanations.

This question got me thinking. 2,500 pounds is 1¼ tons. I can barely budge 90 pounds, and that’s ¹/₁₀ of what the first robot can move. But humans have been lifting massive stone blocks weighing in the tons since ancient times with pure muscle power. How did they do that? They did it by combining strength – as posed by this question – and more subtly by leveraging levers.

Was your answer “yes”, that the robots could lift 2,500 pounds together?

If so, then the simplest reasoning is that the two robots working together combined their lifting powers: 900 + 1,800 = 2,700 pounds that is greater than the boulder’s weight of 2,500 pounds. That reasoning assumes a point weight, i.e., the weight is a tiny point that fits within the joined hands of the robots so that all their lifting power are added directly as diagramed in figure 1.a.

It is a fact that the weight of the boulder can be considered concentrated at a point. In 1993, the Guinness World Records book recognized Gregg Ernst of Novia Scotia for back-lifting 5,340 pounds – that’s almost 2¾ tons! Gregg Ernst managed this feat by positioning himself underneath the exact center of weight of a massive wooden platform that included two cars and their drivers. The entire weight of the platform appeared as a point weight directly under his back, and he only had to concern himself with pushing straight up. It would be impossible for him to actually lift that weight with his hands, but still, what an amazing feat of strength!

What will actually happen with figure 1.a is that both robots will lift half the weight. So, that will overwhelm robot 1 (red Ruby). To successfully lift the point weight, most of the weight must be carried by robot 2 (blue Ben) as in position 1.b. With a point weight, it’s possible for the Ben to lean back enough so that the boulder's center of weight is positioned through its feet for maximum lift.

However, can the boulder really be handled as a point weight that the two robots can scoop up simultaneously within their hands? To answer that question, we need to know how much space 2,500 pounds occupies. For ease of argument, let’s assume the boulder is cube shaped, like those featured in our leader photo. We’ll need to know the density of the boulder because density is the ratio of mass per unit volume: density = mass / volume, or volume = mass / density. This handy Engineering Toolbox page provides a table of densities for many solids. 

Assuming the boulder is common sandstone (140 lbs/ft³), then it’s volume is 2,500 / 140 = 18 ft³ or a cube of dimensions 2.6 ft. That is definitely not a point. To lift the boulder by its opposite edges, the robots would need to be positioned across from each other as in figure 2.a. They would be separated by 2.6 ft and now each would be required to lift exactly half boulder’s weight or 1,250 pounds each. That is beyond the capability of the weaker robot. So, the realistic answer to the problem is no.

You may wonder why each robot must lift 1,250 pounds each. In the figure 2.a (and 2.b), the boulder is being pulled down by a force W directly from its center. That means the robots on opposite side (A and B) must exert forces F₁ and F₂ upwards to counter W. F₁ and F₂ must be equal to lift the weight evenly. Hence, F₁ and F₂ must be at least 1,250 pounds. More specifically, understanding the principal of levers as explained in the video, we can imagine the flat bottom of the boulder as a lever of length 2.6 ft. From the perspective of Ruby, point B is the fulcrum [of a class 2 lever] and it is lifting the entire resistive weight 2,500 with resistive arm of 1.3 ft (half of 2.6 ft) using effort F₁ with effort arm of 2.6 ft. From the principal of levers, we know that F₁ × 2.6 = 2,500 × 1.3. Solving, you find that F₁ = half of 2,500 = 1,250 pounds. Similarly for F₂.

What about if the boulder is composed of the densest earthly material? That would be the very rare metal osmium with density 1,400 lb/ft³ (10-times denser than sandstone). The osmium boulder would occupy 1.8 ft³ or cubic dimension 1.2 ft. That dimension means gripping the boulder at its opposite edges will still create the lever problem as before, making it impossible for the Ruby to lift the boulder.

Alternatively, if Ben wraps its arms around the front and rear facing sides to get closer to the center of weight, then yes, the osmium boulder could be lifted in tandem. The position would be, Ruby lifting by its edge as usual, and Ben's arms positioned sidewise 0.4 ft inward from its edge. Why 0.4 ft? It's an estimate. Since Ben can lift twice the weight of Ruby, if we imagine splitting the bounder vertically into 3 even pieces, left-to-right, then Ruby can lift  ¹/₃ and Ben can lift ²/₃ of the pieces. 1.2 ft ÷ 3 = 0.4 ft. So, Ben's piece will be 0.8 ft wide, placing its center of weight midway at 0.4 ft from Ben, and 0.8 ft from Ruby. 

At Mathnasium, we like to extend knowledge, and oh boy, is this a long post! How then can the robots lift the boulder safely regardless of its size? Why, using same the principal of levers of course! See the next figure. We’ve extended the lever by placing the boulder on a lifting board. The bolder can now be placed off center closer to the stronger Ben. We expect from previous argument that the ratio of the distances d₁ : d₂ = 2 : 1, i.e. if d₂ is 0.4 ft, then d₁ is 0.8 ft. Let’s leverage our knowledge of levers!

Again, Ruby’s perspective is that point B is the fulcrum to a lever of length d₁ + d₂ to counter the weight W at distance d₂. Principal of levers: F₁(d₁ + d₂) = Wd₂. Rearranging d₁/d₂ = (W - F₁) / F₁. This formula gives us the position of the boulder for a given lifting effort of Ruby. If we make Ruby exert a maximum lift of 900 pounds, then d₁/d₂ = 16/9, slightly less than 2/1.

Similarly, Ben’s lever arrangement is F₂(d₁ + d₂) = Wd₁ , and d₁/d₂ = F₂ / (W - F₂). If we make Ben exert a maximum lift of 1,800 pounds, then d₁/d₂ = 18/7, slightly more than 2/1.

Both ratios divide a 25 unit length lifting board. Applying those ratios to the lifting board, we get figure 4.

This figure shows that the weight’s center of weight can be placed within a 2-unit safety window. At the extreme left of that window, 16-units from end A, Ruby will have to exert its maximum lifting force. At the extreme right of that window, 7-units from end B, Ben will have to exert its maximum lifting force. That safety window is important if the robots are not just lifting, but moving that boulder over uneven ground. In that case, we’d like the window to physically accommodate the weight’s center of weight changing direction as the lifting platform tilts up and down. We leave it to the reader to use geometrical modeling to estimate an appropriate physical length. It’s not entirely a theoretical exercise. Human’s have been carrying weights in this manner since ancient times!

We must admit that we’ve had some fun at the expense of the curriculum writer. People often pose simple questions and expect simple answers, but as we’ve shown, interpretation and context can dramatically result in different answers – and endless arguments!

Contact:

Ruby Yao and Benedict Zoe, Mathnasium of Fort Lee
201-969-6284 (WOW-MATH), [email protected]
246 Main St. #A
Fort Lee, NJ 07024

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Photo: //sevenmagicmountains.com/