A small box of mass 4.8kg is sitting on a board of mass 3.6kg and length 1.1m. The board rests on a frictionless horizontal surface. ?
The coefficient of static friction between the board and the box is 0.6. The coefficient of kinetic friction between the board and the box is, as usual, less than 0.6.
Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).
- RealProLv 73 months ago
When you accelerate the board gently what happens? The box moves along with the board.
So there must be an unbalanced horizontal force acting on the box making it move. What force is this? Clearly it's static friction.
If m = 4.8 kg and a = acceleration,
ma = static friction.
What's the maximum possible amount of static friction in this case?
That is defined by the coefficient of static friction.
F_(s,max) = mg * µ _s
If the maximum static friction is mg * µ _s
and m*a is always equal to static friction
Then the maximum value of m*a is also mg * µ _s.
m * a_max = mg * µ _s
a_max = g * µ _s
= g * 0.6
If you try to move the board with a higher acceleration than 0.6g, the box will start to slip off.
Total mass is 8.4 kg.
F_min = 0.6g * 8.4 kg
After the box starts slipping, the friction is kinetic which is even lower than static so the box will slide backwards relative to the board until it falls off. The length of the board is irrelevant unless we have a distance within which we need the board to lose contact with the box.