Max Acceleration of a 20.0kg Crate on a Flatbed Truck

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Homework Help Overview

The discussion revolves around determining the maximum acceleration of a 20.0-kg crate on a flatbed truck, given the coefficient of static friction of 0.400. Participants are exploring the relationship between friction, normal force, and acceleration in a physics context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply the formula for maximum static friction and are questioning how to correctly identify and calculate the normal force. There is confusion regarding the relationship between mass, weight, and the normal force in this scenario.

Discussion Status

Some participants have provided insights into the relationship between weight and normal force, while others are still grappling with the application of the relevant equations. There is a mix of attempts to derive the maximum acceleration, but no consensus has been reached on the correct approach or calculations.

Contextual Notes

Participants express uncertainty about the formulas and concepts involved, indicating a lack of familiarity with the problem type. There is mention of the need to memorize certain equations, and some participants are also exploring related but different scenarios.

pebbles
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Homework Statement


A flatbed truck is carrying a 20.0-kg crate along a level road. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum acceleration that the truck can have if the crate is to stay in place?

Homework Equations


Is the equation to solve for maximum acceleration F s,max= usN * us?


The Attempt at a Solution


I was not taught how to do problems like this...not 100% sure what to plug in for usN...I gave it a shot:
The surface is flat, so I guess the normal force=20.0 kg, so I then mulitplied by .40 to get 8 m/s...ahhhh. I don't know what I'm doing!

Thanks in advance.
 
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pebbles said:

Homework Statement


A flatbed truck is carrying a 20.0-kg crate along a level road. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum acceleration that the truck can have if the crate is to stay in place?

Homework Equations


Is the equation to solve for maximum acceleration F s,max= usN * us?
Fsmax = usN. Then use Newton's 2nd law.

The Attempt at a Solution


I was not taught how to do problems like this...not 100% sure what to plug in for usN...I gave it a shot:
The surface is flat, so I guess the normal force=20.0 kg, so I then mulitplied by .40 to get 8 m/s...ahhhh. I don't know what I'm doing!
20 Kg, is the mass, the normal force is its weight. Use formula that relates weight with mass.
 
PhanthomJay said:
20 Kg, is the mass, the normal force is its weight. Use formula that relates weight with mass.
uh, hmm. I looked through my notes, couldn't find this kind of formula.

What am I to do to solve this problem? I looked at the F s, max formula wrong, so it's us times N. I was initially puzzled by usN times us. Now it makes since.

So, us= .40 right? But I don't know how to get the weight of 20kg for normal force, N.

Thanks
 
The normal force on the crate, in this problem, is equal to the weight, since there is no acceleration vertically and only the weight and normal forces act in the vertical diretion, so they must be equal and opposite in satisfaction of Newton's 1st law. Now weight is just the force caused by the acceleration of gravity, please memorize it, W=mg, where g is the acceleration of gravity where g is equal to 9.8m/s/s on the Earth's surface. When the mass is in kilograms, the weight, using this formula, is in Newtons. 20 Kilograms weighs 196 Newtons.
 
so my answer will be .40 times 196 N which is 78 m/s^2?!
 


A flatbed truck is carrying a 20.0-kg crate along a level road. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum acceleration that the truck can have if the crate is to stay in place?

f=ma for (m) moving direction
R=ma
F=mg
coeffi.fric=F/R=mg/ma
a=g/coeff.fri
= 10/0.4m/s2
=25m/s2
 
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I love this problem and since I am a flatbed driver I have a more interesting one:

48 feet flatbed
6000 pounds load 8 feet high (the load is full trailer)
A flatbed is 8.5 feet whide
driver will drive at 65 miles per hour (there is no wind)

How many straps and how tiht should they be so he does not lose everything in 600 miles :D

(I'm a geek flatbed driver... If I get an answer to this I will post more)
 
Last edited:

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