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Question
(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 by OpenStax is licensed under CC BY 4.0.
Final Answer

a) $4.9 \textrm{ m/s}^2$

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 mg 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 ma equals mu mg over two. We’ll cancel the ms on both sides and we solve for a 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 a max by dividing this by m and this by m, and the ms cancel giving us mu s g 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 mg. So we have the maximum static friction force is mu s times mg, and the ms cancel because you have ma 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 mu s g 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.

Comments

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???