What P Value Causes Sliding or Tipping of a Box?

[NGamer]

A box is 8 ft high, 4 ft deep, and 6 ft wide and weighs 270 lb. A worker pushes on the box at a location along the width and 5 ft above the floor. The coefficient of friction between the box and the floor is .53. 1) Determine what value of the pushing force, P, will cause sliding. 2) Determine what value of the pushing force, P, will cause tipping.

This is an ungraded question for class. I do not have a physics background so am unclear how to solve.

 

[AdityaDev]

In this situation, according to the magnitude of applied force, the normal reaction force shifts towards the point about which the box has a tendency to rotate so as to balance the torque applied by the external force (how?...The wieght that still acts vertically downwards at centre of mass will start opposing the rotation. If Normal reaction stayed there, it would have canceled the force Mg) Even if the normal shift, from free body diagram, $N=Mg$.
Along the horizontal direction,$F-\mu Mg=Ma$
If you apply force on the left side of the box, the box will try to rotate about the bottom right corner and so you can apply the torque equation as well about that point.
you now have enough info for solving first part.

for second part, the normal will try hard to stop rotation and will move to the edge. That will be the limitting case. Normal can't move out of the box. Now the forces that can create torque is mg and F (Normal and friction act along that point).
So if the edge of cube is a and b is the distance from corner to the point where F is applied, this equation gives the value of F.
$$Fb=mg\frac{a}{2} $$

 

[NGamer]

While I appreciate your insight, I am still very confused - esp based on barrel examples that I've reviewed.

So are you advising to complete the formula, F - μMg = Ma...if so, I'm still unclear...like: F - 0.53(270 lb)(32.2 ft/s^2) = ?

 

[haruspex]

So are you advising to complete the formula, F - μMg = Ma...if so, I'm still unclear...like: F - 0.53(270 lb)(32.2 ft/s^2) = ?

You don't need any substantial acceleration, the tiniest will do. So you can set the acceleration to zero.
For the second part, draw a free body diagram. (If you don't know how to that, read up on it. It is an essential skill.) Assume it is just starting to tip.

By the way, this belongs in Introductory Physics, not Advanced.

 

[NGamer]

Thanks! Apologies for it being in the wrong section...like I said, no physics experience, but learning.

Again, thanks for the assistance.