What Is the Maximum Acceleration for a Truck Carrying a Crate?

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The discussion revolves around calculating the maximum acceleration of a truck carrying a 60.0 kg crate on a level road, given a coefficient of static friction of 0.490. The correct approach involves using the formula for static friction, where the maximum static friction force equals the coefficient of static friction multiplied by the normal force. After applying the formula, the maximum acceleration is determined to be 4.81 m/s². Participants clarify the importance of focusing on the coefficient of static friction rather than kinetic friction. The conversation concludes with a successful solution and tips for future calculations.
kaylanp01
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I'm sure that this is a very simple question, but I can't find it in my text. I made a few attempts but I see why they're incorrect, I just can't figure out the correct method.

Homework Statement


A flatbed truck is carrying a 60.0 kg crate along a level road. The coefficient of static friction between the crate and the bed is 0.490. What is the magnitude of the maximum acceleration that the truck can have if the crate is to stay in place?

Homework Equations


I used N-mg=may.

The Attempt at a Solution


N=mg
0.490/60 = 0.008 ( I see why this is incorrect).

and

N=mg=(60)(-9.81)=-588.6 N (I think that I have to use this somehow).

Thanks!
 
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Hi kaylanp01! :smile:

(have a mu: µ :smile:)

Hint: ignore m … the m's all cancel … the acceleration is the same for any object with the same µ. :wink:
 
tiny-tim said:
Hi kaylanp01! :smile:

(have a mu: µ :smile:)

Hint: ignore m … the m's all cancel … the acceleration is the same for any object with the same µ. :wink:

I still don't understand. Do I need to introduce a new formula? Or do I have everything that I need there already?

Thanks for the µ :).
 
Hi kaylanp01! :smile:
kaylanp01 said:
I still don't understand. Do I need to introduce a new formula? Or do I have everything that I need there already?

Thanks for the µ :).

erm … well, you need a formula involving µ, for a start! :redface:
 
tiny-tim said:
Hi kaylanp01! :smile:


erm … well, you need a formula involving µ, for a start! :redface:


haha hey now, be nice. its a saturday night and I'm doing a physics capa, how good is my mindset supposed to be right now? :P

the only formula that I can find with µ is the one for kinetic friction, but i can't see how that's significant. am i missing something?
 
kaylanp01 said:
haha hey now, be nice. its a saturday night and I'm doing a physics capa, how good is my mindset supposed to be right now? :P

wot's a capa? :confused:
the only formula that I can find with µ is the one for kinetic friction, but i can't see how that's significant. am i missing something?

mmm … how about an equation relating normal force to static friction? :wink:

(or, to put it another way, what is the coefficient of friction?)
 
tiny-tim said:
wot's a capa? :confused:


mmm … how about an equation relating normal force to static friction? :wink:

(or, to put it another way, what is the coefficient of friction?)

a capa is hell in the form of an online assignment. haha.

okay, i have the formula that you're talking about, i think...and i assume that it gives you
fs(max)= µsN=0.490(-588.6)=288.4

fs=ma
-288.4=(60)a
a=-4.81 m/s^2=4.81 m/s^2

now i shall insert that answer andddd...
correct!

thank you thank you! :)
 
kaylanp01 said:
okay, i have the formula that you're talking about, i think...and i assume that it gives you
fs(max)= µsN=0.490(-588.6)=288.4

fs=ma
-288.4=(60)a
a=-4.81 m/s^2=4.81 m/s^2

now i shall insert that answer andddd...
correct!

:biggrin: Woohoo! :biggrin:

… though you would get even more marks if you didn't put those m's in. :wink:
 
tiny-tim said:
:biggrin: Woohoo! :biggrin:

… though you would get even more marks if you didn't put those m's in. :wink:

the sad part about a capa is that every question is only worth 1 mark :(, but i will keep that in mind for my exam :).
 

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