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(a) If half of the weight of a small $1.00 \times 10^3 \textrm{ kg}$ utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both problems assuming the truck has four-wheel drive.
Question Image
<b>Table 5.1</b> Coefficients of Static and Kinetic Friction
Table 5.1 Coefficients of Static and Kinetic Friction
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
  1. $4.9 \textrm{ m/s}^2$
  2. $4.9 \textrm{m/s}^2$
    Note: In the video for part b) I made the mistake with the coefficient of static friction on the wrong assumption that the truck box was made of metal. The coefficient of static friction should be that for metal on wood, which is 0.5.
  3. $9.8 \textrm{ m/s}^2$
Solution Video

OpenStax College Physics Solution, Chapter 5, Problem 5 (Problems & Exercises) (2:48)

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Video Transcript

This is College Physics Answers with Shaun Dychko. Let’s assume this truck is a front-wheel drive truck as most trucks are and we’re told that the normal force between these two driving wheels is half of the weight. So in our formula for this maximum static friction force, this normal force is going to be <i>mg</i> over two, the weight of the truck divided by two. This is also the net force as it turns out because there is only one force horizontally and that’ll be this friction force. So this static friction force is going to equal mass times acceleration, and so we can say <i>ma</i> equals <i>mu mg</i> over two. We’ll cancel the <i>m</i>s on both sides and we solve for <i>a</i> then. And that’s the static friction between rubber and dry concrete which in Table 5.1 is 1.0, we multiply that by 9.8 meters per second squared and divide by two to get 4.9 meters per second squared will be the acceleration. And the metal filing cabinet in the flatbed of the truck, it’ll have a maximum static friction force of the coefficient of static friction between metal on metal, or steel on steel, and multiplied by the weight of the filing cabinet, and that’s going to equal mass times the maximum acceleration that it could possibly withstand without sliding. So we solve for <i>a max</i> by dividing this by <i>m</i> and this by <i>m</i>, and the <i>m</i>s cancel giving us <i>mu s g</i> and that is 0.6 when you look it up on Table 5.1, times 9.8, which is 5.9 meters per second squared. So since the actual acceleration of 4.9 is less than the maximum that’s possible, then the cabinet will not slip. Then in part c, now since we have four-wheel drive, the entire weight of the truck is being supported by the driving wheels since all four wheels are driving and so the normal force now is <i>mg</i>. So we have the maximum static friction force is <i>mu s</i> times <i>mg</i>, and the <i>m</i>s cancel because you have <i>ma</i> for the net force as well. I’ve skipped a bit of algebra here but it’s the same as for part a, so the acceleration is <i>mu s g</i> so it’s one times 9.8, which is 9.8 meters per second squared. Now in this case, the cabinet will slip since the actual acceleration is greater than the maximum possible acceleration given the coefficient of static friction between steel and steel.


Submitted by I love Physics on Sun, 11/18/2018 - 20:26

I don't get why in Number b, you didn't divided the g(9.81) by 2???

Submitted by ShaunDychko on Tue, 04/02/2019 - 12:13

Thanks very much for your question, and sorry I didn't get back to it sooner. It might not be relevant to your course anymore, but hopefully is to other students!
In part a) we divided the truck weight by 2 since the front tires carry half the truck weight (the question tells us this). In part b) on the other hand, we're dealing with a metal cabinet in the truck box. The normal force applied by the truck box on the metal cabinet is the full weight of the cabinet, hence no need to divide by 2. The weight of the cabinet is being supported by only one thing, namely the truck box, whereas the weight of the truck is being supported by 2 sets of wheels, only one of which is applying static friction to the ground since we assume it's a front wheel drive truck.

In reply to by I love Physics

Submitted by Devin.Reid on Fri, 03/29/2019 - 17:24

Just to clarify. The question said "metal cabinet lying on the [wooden] bed of the truck." So, the coefficient in part b should be 0.5 not 0.6.

Submitted by ShaunDychko on Tue, 04/02/2019 - 12:14

Thank you Devin.Reid, you're quite right. I made the mistake of assuming the truck box was metal. I have put a note in the Final Answer section about this.

In reply to by Devin.Reid

Submitted by aspruce on Wed, 01/01/2020 - 15:35

oof. this video should be re-done. your mistake makes it too hard to follow for the answers.

Submitted by leahojog on Sat, 02/22/2020 - 12:49

Why do you use the formula for static friction? If it's accelerating (ie., moving), shouldn't it be the kinetic friction formula?

Submitted by jdanner917 on Thu, 03/12/2020 - 16:02

How do you say that 4.9 m/s2 is the maximum acceleration for the truck? We don't know the force that the truck is putting down with the tires. you just gave the force needed to get the truck moving, right?

Submitted by ShaunDychko on Mon, 09/21/2020 - 17:00

Thanks for the question jdanner917. The force to accelerate the truck is due to static friction, and there's a limit to how large that static friction force can be. In part (a) we see that the force that the truck is putting down on with the tires is $mg/2$ on each tire since it's being pulled down by gravity, and the ground therefore pushes up on the tires with the same force since the truck isn't accelerating vertically. We know the force on the tires, in other words, but not as a number (not in newtons). That's OK, though, since the mass of the truck cancels on both sides of the equation since the friction force is the net force and we can therefore set it equal to $ma$. Since there are two tires driving the truck (not 4 in this case), and we're presuming the coefficient of friction of rubber on dry concrete of $1.0$, the video shows dividing $g$ by 2 (since there are two tires) to arrive at 4.9 m/s^2.
Hope this helps,

In reply to by jdanner917