# Weight distribution's effect on riding uphill

Two different chains of cars (attached to each in a row) are driving uphill (Only the first car is motorized): The first chain consists of 3 cars, each 4000 kg. The second chain consists of 3 cars, the most front one being 8000kg and the others 2000kg each. Thus, meaning that the total weight is equal for both chains, but the weight is distributed differently.

For which chain will it be the easiest to drive uphill?

Somehow I think it has something to do with the c.g. or c.o.m or friction.. but i can't come up to a logical explanation.

Note: I received the questions exactly as I present them here from the organization I got them from. Therefore, I suffer from not being able to provide more context and specifications. I am only being able to provide my assumptions based on my preliminary knowledge.

Thanks in advance for the assistance!

The FBD I draw for the "8000,4000,4000" chain

• Is only the first car motorized, like a train? If so, since the force on the wheels due to friction, that allows the motorized first car to pull the rest up, which depends on the normal force, which in turn depends on the mass of only the first car only could be "easier" in that sense. That's a bit sloppily worded, but I think it gets the gist across. Sep 2, 2018 at 0:35
• Yes, exactly, only the first car is motorized. Sep 2, 2018 at 0:38
• No problem, I think I got what you meant: So, the normal force acting on the, motorized, first car defines how easy the car chain will go up; so, the greater the normal force (which is of course the case when the weight distribution moves towards the motorized car), the easier to pull up? Sep 2, 2018 at 0:45
• In which case, moving weight from the rear cars to the first car increases the normal force available for allowing the motor in the first car to move the weight up the hill while simultaneously reducing the weight in the rear cars that "tries" to pull the first car back down the hill. Draw a free body diagram for the first car and I think you'll see that the required coefficient of static friction for the heavy first car case is lower than the case for the equal weight cars. Sep 2, 2018 at 0:47

Call the total mass of the rear cars $m_2$ and the mass of the first car $m_1$. Along the slope of the hill the forces on the first car are

$$F = m_1 a = \mu m_1 g \cos(\theta) - m_2 g \sin(\theta)$$

To keep the cars desired direction when the train of cars is not accelerating ($a=0$) the coefficient of friction must be at least

$$\mu \ge \frac{m_2}{m_1} \tan(\theta).$$

From this perspective, anything you can do to make $m_2$ smaller and $m_1$ larger will make the trip easier.

• Awesome, thanks a lot for your explanation. The first thing I am going to do tomorrow is to draw the FBD and use the info you provided to proof the theory. I will couple back my results, in the hope that you could throw a quick look over it. Sep 2, 2018 at 1:03
• Alright, I have drawn the FBDs for the (8000, 2000, 2000 chain). I am just a freshman, so for verification: μ is the "k" in "Ffriction = k x Fnorma"l, and "m2/m1 tan(θ)" is the formula with which the friction coefficient "k" can be calculated. Did I get that right?If so, I, indeed, arrive at a Fres of zero, and μ is roughly 0,28, whereas it is 1,15 for the 4000-4000-4000 chain. So, the smaller μ, the easier the trip uphill, right? Sep 2, 2018 at 20:47
• That is my interpretation of the problem. Yes. Whether that is the answer the professor is looking for is up to you and the context the problem was introduced. Sep 3, 2018 at 16:02

As stated the question is pretty vague. Are you sure this is about friction and not just about the forces (tensions) in the chains between the cars? Yes for sure the first car needs traction (friction) but really since both trains have the same total mass both go up equally as well.