# Why do objects with greater length feel heavier - and how to calculate perceived weight?

Consider this situation:

As part of some training, you are asked to pull an object that is 100 feet long, weighing approximately 218 pounds across a distance of 310 feet (fire hose across concrete if materials matter). While carrying a 218 pound object that far would be difficult no matter what the method is, it seems to feel a lot heavier when the object is spread out over 100 feet. I know there is some calculations for torque that can be done, but none of the numbers I'm getting seem to make any sense. Is there a way to semi-accurately calculate how much "effort" it would take to move an object of this size? Does the disbursement of the weight actually make a difference? I know there could be a lot of factors here such as friction, the angle of dragging the hose, possibly the speed at which the hose is moved, etc. I've included some explicit and implicit values that may be of use.

• Hose Length: 100 feet
• Hose Weight: 218 pounds
• Distance: 310 feet
• Pull Speed: ~1mph
• Average Height of puller: ~70 inches
• Distance between feet and where the hose touches the ground when carried: ~5 feet

Hopefully this question makes sense. I may be over thinking it. Please let me know if you need any more information.

There is indeed an accurate way to calculate the amount of what you call "effort". In physics, this concept is known as work $$W$$, and is the energy transferred to an object by letting a force $$\vec F$$ act on it along a path $$s$$. Mathematically, this is written as $$W = \int_{s_0}^{s_1}\vec F\cdot d\vec s$$ where $$s_0$$ and $$s_1$$ are the beginning and endpoint of the path, respectively. If the magnitude $$F$$ of the force is constant along the path (as is reasonable to assume in your case), this reduces to the simpler form $$W = F(s_1-s_0)$$ where now $$s_1-s_0$$ is simply the length of the path.

Let us now analyze your specific example. The total force acting on the hose is composed of the force required to lift the end you are holding and the friction between the hose and the ground. Note that these two components are orthogonal: One acts purely vertically, while the other acts purely horizontally. A great advantage of the two forces being orthogonal is that we can write $$F$$ in terms of its components easily by using Pythagoras' theorem: $$F = \sqrt{F^2_\text{lift}+F^2_\text{friction}}$$ The length of the hose only affects the horizontal ("friction") component, which is in turn proportional to the area $$A$$ of contact between the hose and the ground: $$F_\text{friction} = \kappa A = \kappa d l$$ Where $$\kappa$$ is a constant. Technically, the friction also depends on the speed and the weight, but since we assume these to be constant, this dependence can be absorbed into $$\kappa$$. The area is approximately determined by the diameter $$d$$ of the hose times the length $$l$$ that actually touches the ground, which is just the total length $$100\text{ft}$$ minus the length which is held up, again determined by Pythagoras' theorem to be $$l = 100\text{ft}-\sqrt{(70\text{in})^2+(5\text{ft})^2}$$
Plugging all of this into our original formula for the work we get $$W = 310\text{ft}\cdot\sqrt{F^2_\text{lift}+(\kappa d l)^2}$$ and see that the force depends linearly on the length of the hose. Torque is not needed in this case, becsuse there is no rotational motion involved.

• Wow, this answer is definitely more complex that I expected. Fantastic explanation! This makes a lot of sense, but I didn't understand a couple of areas. I'm a computer science major so please forgive the ignorance on my part. How would I determine the value of the constant 'k' here? Commented Jun 24 at 7:01
• Unfortunately, such constants can only be experimentally determined. One way to do that would be to put a newtonmeter between yourself and the hose, and then going back through the formulae I provided to calculate it. Commented Jun 24 at 15:05

I expect a long hose is harder to drag because most of the length is pressing on the concrete. There is a lot of friction.

You could pick up a shorter object, eliminating the friction.

@Paulina mostly has it right (+1). I will add this.

$$W = \vec F \cdot \vec d = (\vec F_{lift} + \vec F_{drag}) \cdot \vec d$$

$$\vec F_{lift}$$ is vertical, while $$\vec F_{drag}$$ and $$\vec d$$ are horizontal and in the same direction. So

$$W = \vec F_{drag} \cdot \vec d = F_{drag} d$$