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A box on a truck tipping

  • Thread starter Zach981
  • Start date
  • #1
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Homework Statement



A truck is loaded with a large box, which is a uniform rectangular solid 1.8 m tall, 1.0 m wide, and 1.2 m deep.
The box sits upright on a truck with its 1.0 m dimension in the direction of travel, and the bed of the truck is sufficiently rough that the load cannot slide. How rapidly can the truck accelerate without tipping the box over? [Hint: Suppose the box is just starting to tip; where is the normal force acting?]


Homework Equations



This is from the torque/equilibruim section of the physics textbook.

The Attempt at a Solution



I set up a free body diagram, made three equations for the forces in the x direction, y direction, and the net torque. But mastering physics says my answer is wrong. How should I set up this problem?
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Hi Zach981! :wink:
I set up a free body diagram, made three equations for the forces in the x direction, y direction, and the net torque.
Show us your equations. :smile:
 
  • #3
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hi zach981! :wink:


Show us your equations. :smile:
4567.png

What did I do wrong?
 
  • #4
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I have 2 hours left to finish this assignment, does anyone here have know what to do?
 
  • #5
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Forget it, figured this out myself,
μ = cosθ/sinθ
where:
tanθ = 1.8 m/ 1 m

then, a = μg

just in case anyone else has to do these stupid mastering physics assignments.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi Zach981! :smile:

(just got up :zzz:)

(in future, show your working at the start, as per the forum rules, and you'll receive help much sooner)
View attachment 52411

What did I do wrong?
Forget it, figured this out myself,
μ = cosθ/sinθ
where:
tanθ = 1.8 m/ 1 m

then, a = μg

just in case anyone else has to do these stupid mastering physics assignments.
sorry, but both solutions are the wrong idea

moment of inertia has nothing to do with it … that would only matter if you were interested in the rate of turning

the coefficient of friction, µ, also has nothing to do with it … the tipping angle would be the same even if the box had a fixed pivot at the corner, wouldn't it? :wink:

your answer (a = gtanθ) is correct, but you need to rewrite your reasoning :smile:

(btw, you could achieve the same result more easily, if your professor allows you to use non-inertial frames, by adding a horizontal "fictitious" force of -ma :wink:)
 

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