# Sliding and toppling.

Gold Member
Ok, I have a maths question. This is a little different because I dont want the answer, I want the formula so I can work out the answer (and future answers) myself. lol it's not very good self-learning with a book that doesnt tell you much. Anyway here is the question.

A solid uniform cube of side 4cm and weight 60N is situated on a rough horizontal plane. The coefficient of friction between the cube and the plane is 0.4. A force, P, acts in the middle of one of the edges of the top of the cube, at right angles to it and at angle theta to the upward vertical.

In the cses when the value theta is (a) 60 degrees (b) 80 degrees, find

(i) the force P needed to make the cube slide, assuming it does not topple;
(ii) the force P needed to make the cube topple, assuming it does not slide.

So I have tried many variations using sin, cos and tan but cannot seem to get the right answer. Can someone tell me the formula so i can work these out for myself?

Doc Al
Mentor
Don't think in terms of finding "the formula". Instead, learn the basic physics involved and then you can figure out all the formulas you need no matter what.

As always in these kinds of problems, consider all the forces acting on the object. I count four: (1) the applied force P, which acts along the angle you described (θ from vertical), (2) the weight of the cube, (3) the friction of the plane against the cube, and (4) the normal force of the plane up on the cube.

Draw a picture!
jimmy p said:
(i) the force P needed to make the cube slide, assuming it does not topple;
In this case you need to overcome the static friction to get the cube to slide. Consider the limiting case where the force P is as big as it can be without causing translation. The cube is in equilibrium:

The sum of the forces in the y-direction must equal zero, so:
P cosθ -mg + N = 0
The sum of the forces in the x-direction must also equal zero, so:
P sinθ -μN = 0

Use these equations to figure out P, given θ.
(ii) the force P needed to make the cube topple, assuming it does not slide.
In this case we want it to topple (tip over) not translate. So this time we consider torques about a pivot point. Consider the case where the force P just barely causes the cube to start to tip. At that point the torques about the edge must equal zero, so:
mgL/2 - P cosθ(L/2) - P sinθL = 0
("L" is the length of the side of the cube.) I'll leave it for you to figure out what I did to get that equation. 