How do I determine the tipping angle of a box on a truck?

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To determine the tipping angle of a box on a truck, one must analyze the forces acting on the box, particularly the normal force when the box is about to tip. The correct approach involves using the tangent of the angle formed by the box's height and width to find the acceleration limit before tipping occurs. The equations for forces in both the x and y directions, along with net torque, should be set up properly to solve the problem. The solution involves recognizing that the acceleration can be calculated using the relationship a = g * tan(θ), where θ is derived from the box's dimensions. Understanding these principles is crucial for solving similar physics problems effectively.
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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?
 
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Hi Zach981! :wink:
Zach981 said:
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:
 
tiny-tim said:
hi zach981! :wink:


Show us your equations. :smile:

4567.png


What did I do wrong?
 
I have 2 hours left to finish this assignment, does anyone here have know what to do?
 
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.
 
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)
Zach981 said:
View attachment 52411

What did I do wrong?
Zach981 said:
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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