- $4.9 \textrm{ m/s}^2$
- $4.9 \textrm{m/s}^2$
**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. - $9.8 \textrm{ m/s}^2$

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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.

## Comments

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

Submitted by shelby.keith on Fri, 03/01/2019 - 14:51

In reply to I don't get why in Number b,… by I love Physics

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

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 I don't get why in Number b,… by I love Physics

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

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

In reply to Just to clarify. The… by Devin.Reid

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

Submitted by ShaunDychko on Wed, 01/01/2020 - 20:07

In reply to oof. this video should be re… by aspruce

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

Submitted by mmh1914 on Sat, 09/19/2020 - 21:22

In reply to Why do you use the formula… by leahojog

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

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

Hope this helps,

Shaun

In reply to How do you say that 4.9 m/s2… by jdanner917