Calculating force to tip over a cabinet (static equilibrium)

In summary, the conversation is about a physics problem involving finding the force, coefficient of static friction, and minimum force required to tip over a cabinet. The person attempting to solve the problem tried setting up tables for forces and torques but was unsure how to relate them. They were also unsure of the meaning of "tipping" and the type of motion involved. Through the conversation, they come to the conclusion that a net torque is needed to cause a change in the rotational state of motion, and they attempt to find expressions for the torques exerted on the block. However, there is some confusion about the angle and the torque generated by the weight.
  • #1
Mitch17

Homework Statement


physics chapter 12 homework number 4.PNG


Homework Equations


How do you solve for the force in this problem and also the coefficient of static friction and minimum force to tip the cabinet over if the point of application can be chosen anywhere on the cabinet?

The Attempt at a Solution


I tried setting up tables that represent the forces acting on the object in the x and y direction and also tried setting up a table for torques, but I could not figure out how to relate the torque to the force in the problem and solve for the force
 

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  • #2
Can you do part (a) first? The point of application of the external force is given and you may assume that the coefficient of static friction is large enough so that the block will tip before it starts sliding.

On edit: If you need help for parts (b) and (c), please provide a better picture or type in the questions.
 
  • #3
@kuruman No not really, I basically have no idea where to start.
 
  • #4
What does "tipping" mean? Can you describe in your own words using ideas from physics what happens when a block tips?

On edit: More specifically, can you describe the kind of motion that a block undergoes when it tips?
 
  • #5
@kuruman When the block starts to tip, I think the force of static friction is still acting on the block because it would pivot around the bottom right corner and would not slide in the x direction. I think when the block starts to tip that the force of gravity on the block would cause a torque and the block would start to accelerate in the negative y direction toward the ground. The force F that is pushing the block also causes a torque and that is how the block starts to move in the first place.
 
  • #6
The key words in what you wrote is
Mitch17 said:
... it would pivot around the bottom right corner ...
In other words, the block will rotate about the right corner. OK, so what is needed to cause an object to change its rotational state of motion?
 
  • #7
A net torque causes a change in its rotational state of motion.
 
  • #8
Correct. Draw a free body diagram of the block. How many different forces exert torques on this block about the pivot point? Can you find expressions for these torques?
 
  • #9
I found the expressions τF = F(1m)sin(-θ) and τmg = mg(0.5m)sin(-90°)
 
  • #10
What do you think is angle θ? Ignore the drawing and read the statement of the problem carefully.
 
  • #11
The angle θ is 0 if since the force is applied horizontally, so my equation for torque would be τF = F(1m)sin(-90°).
 
  • #12
If θ is 0, why did you set θ = 90o in your expression τF = F(1m)sin(-θ)? Why the negative sign anyway? It looks like you are a bit confused about what θ means in the expression for the torque. Be sure to sort that out. Also, recheck the torque generated by the weight. There is something incorrect about it.

It's way past my bedtime where I am, so I have to quit for a few hours.
 

Related to Calculating force to tip over a cabinet (static equilibrium)

What is static equilibrium?

Static equilibrium is a state where an object is at rest and all forces acting on it are balanced, resulting in no net movement.

What factors affect the force needed to tip over a cabinet?

The force needed to tip over a cabinet depends on its weight, height, and the distance of its center of gravity from its base. The weight and height determine the magnitude of the force, while the distance of the center of gravity from the base affects the stability of the cabinet.

How do you calculate the force needed to tip over a cabinet?

The force needed to tip over a cabinet can be calculated using the formula F = mgd, where F is the force, m is the mass of the cabinet, g is the acceleration due to gravity, and d is the distance of the center of gravity from the base of the cabinet.

What are the units of force and how do they relate to tipping over a cabinet?

The unit of force is Newton (N), and it is the product of mass (kg) and acceleration (m/s^2). In the context of tipping over a cabinet, the force needed is directly proportional to the mass and the distance of the center of gravity from the base. Therefore, the greater the mass or the further the center of gravity is from the base, the greater the force needed to tip over the cabinet.

How do you ensure static equilibrium when designing a cabinet?

To ensure static equilibrium, the center of gravity of the cabinet must be located within the base of the cabinet. This can be achieved by evenly distributing the weight of the cabinet and placing heavier items closer to the bottom. Additionally, the base of the cabinet should be wide and stable enough to support the weight and prevent tipping over.

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