What Forces Prevent a Refrigerator from Tipping on a Moving Truck?

[Trentonx]

Homework Statement

A refrigerator is approximately a uniform parallelepiped h = 8 ft tall, w = 3 ft wide, and d = 2 ft deep. It sits upright on a truck with its 3 ft dimension in the direction of travel. Assume that the refrigerator cannot slide on the truck and that its mass is 130 kg. For the first three parts of this problem, the rope shown in the picture is not there.
c) What is the maximum acceleration the truck can have such that the refrigerator does not tip over?
d) Suppose now that a rope connects the top of the refrigerator with the cab of the truck, which now accelerates at twice the acceleration calculated in (c). The refrigerator lifts off slightly at the front but is held in place by the horizontal rope. Find the tension in the rope.

Homework Equations

F=ma
T=Ia
T-rFsin(theta)

The Attempt at a Solution

So I guess that the frig is just starting to tip, so only the back corner is touching. That's the pivot point then, but I'm lost about the forces and where they are acting
I know gravity acts on the center of mass, a normal force on the contact point, but there must be more.

 

[LowlyPion]

First figure where the center of gravity is.

The say the density is apparently uniformly distributed, so it's a dimensional game then isn't it?

The center of gravity is acting down and it is being accelerated forward, so the center of gravity is a force moving back.

What are the sum of the moments of the forces? When they are just equal you are at the tipping point.

 

[Trentonx]

So the center of mass would be 4ft up, 1.5ft over and 1ft back. The center of mass is the same as center of gravity, right, so that's where the mg acts. I don't see how the center of gravity becomes a force moving back though.
The sum of the torques must be zero so that the fridge doesn't go all the way over. The contact forces act at the pivot point, so they don't matter, and gravity is keeping the fridge from tipping, what's making it tip?

 

[LowlyPion]

So the center of mass would be 4ft up, 1.5ft over and 1ft back. The center of mass is the same as center of gravity, right, so that's where the mg acts. I don't see how the center of gravity becomes a force moving back though.
The sum of the torques must be zero so that the fridge doesn't go all the way over. The contact forces act at the pivot point, so they don't matter, and gravity is keeping the fridge from tipping, what's making it tip?

The truck is accelerating the icebox. Consider its inertia. The acceleration of the truck means the CoM wants to lag horizontally backwards.

Moreover, when the box begins to lift the CoM rises, making the moment arm longer for the horizontal force and the vertical m*g acts slightly closer to the back pivot, meaning once it starts, it goes.

 

[Trentonx]

So the refrigerator resists the change, and the truck bed exerts a force, so the refrigerator tries to tip backward. As it tips, then where mg acts also changes, and contributes to the torque.
So how do I start accounting for this in equations? Especially the part about the moment arm increasing in length?
I know
\sumt=0

 

[LowlyPion]

I would set them equal in the equation, and be confident that once it gets to be equal it will begin to tip and keep on tipping.

So figure the moments then. Let's put some numbers for what you're dealing with.

 

[kermitthefrog]

so if I understand this correct then m*g*r*H=m*a
where m is the mass of the fridge, g is acceleration due to gravity, r is the radius it rotates around =4m? , H is the height of the fridge=8 meters, and a is the acceleration of the truck

This gives a huge acceleration, which can't be right. Anyone else have any ideas on what is missing?

 

[Redbelly98]

Welcome to Physics Forums.

so if I understand this correct then m*g*r*H=m*a

Since the units are different on the two sides of that equation (N*m² vs. N), it cannot be right.

If you can show your work, we can look at where the mistake is. Also, did you draw a free-body diagram? It would help if you list the forces you used, or upload an image of the diagram.

 

[vigintitres]

You'll need the moment of inertia for this, that's for sure. Probably have to use the parallel axis theorem as well.