Can You Move a Heavy Block Without Tipping It Over?

[KLscilevothma]

Physics contest -- question 2

Suppose you are to move a piece of heavy furniture at home. Let us approximate it by a cubic block of unifor mass M on a rough floor surface with friction coefficient u=1. Your strength can only provide a force F=0.8Mg, where g is the gravity acceleration, so you cannot simply lift the block and move it around.

a) If you apply the force as shown in the figure, what will happen to the block?

b) You are free to apply the force F in any direction to any point of the block, how to apply the force so the block will have maximun acceleration in sliding motion but without tipping over?
(I know part b is pretty easy but I failed to answer it )

[There may be many ways and you are only required to find one. YOu many also need this : sinθ+cosθ=1.414sin(θ+45)]
[Hint: The magnitude of the friction is the smaller of the two : F or uN, where N is the reaction force of the floor perpendicular to the furface. The effect of the friction is to keep the block from sliding until it reachers its maximum value uN)

a) It will tip over. (I answered it in more detail during competition)

 

[AndyW]

b) To achieve maximum acceleration in sliding motion without tipping over, the force F should be applied at a point on the block that is at a distance of 0.414 times the block's height from the bottom edge, at an angle of 45 degrees from the horizontal. This will result in a net force of 0.8Mg acting in the direction of motion, while also creating a moment arm of 0.414 times the block's height which will counteract the tipping motion.

To understand this, let's break down the forces acting on the block. The force F applied by the person can be resolved into two components - one parallel to the floor, which is responsible for the block's acceleration, and one perpendicular to the floor, which creates a moment arm and causes the block to tip over.

The maximum acceleration in sliding motion can be achieved when the parallel component of F is equal to the frictional force acting on the block. This means that F must be equal or greater than uN, where N is the normal reaction force from the floor.

Now, using the given hint, we know that the magnitude of the frictional force is the smaller of the two forces - F or uN. This means that in order to achieve maximum acceleration, the force F must be equal to uN.

To prevent the block from tipping over, the perpendicular component of F must be counteracted by an equal and opposite moment. This can be achieved by placing the force F at a distance of 0.414 times the block's height from the bottom edge, at an angle of 45 degrees from the horizontal. This creates a moment arm that is equal to the perpendicular component of F, effectively balancing out the tipping motion.

In summary, by applying the force F at a distance of 0.414 times the block's height from the bottom edge, at an angle of 45 degrees from the horizontal, we can achieve maximum acceleration in sliding motion without tipping over the block.