Calculating force to tip over a cabinet (static equilibrium)

[Mitch17]

Homework Statement

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

 

[kuruman]

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.

 

[Mitch17]

@kuruman No not really, I basically have no idea where to start.

 

[kuruman]

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?

 

[Mitch17]

@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.

 

[kuruman]

The key words in what you wrote is

... 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?

 

[Mitch17]

A net torque causes a change in its rotational state of motion.

 

[kuruman]

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?

 

[Mitch17]

I found the expressions τ_F = F(1m)sin(-θ) and τ_mg = mg(0.5m)sin(-90°)

 

[kuruman]

What do you think is angle θ? Ignore the drawing and read the statement of the problem carefully.

 

[Mitch17]

The angle θ is 0 if since the force is applied horizontally, so my equation for torque would be τ_F = F(1m)sin(-90°).

 

[kuruman]

If θ is 0, why did you set θ = 90^o in your expression τ_F = 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.