# Distance to Come to Rest and Force of Friction

1. Sep 26, 2015

### Koru

1. The problem statement, all variables and given/known data
The coefficient of static friction is 0.625 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.137. Force F causes both blocks to cross a distance of 4.46m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.57kg and the mass of the upper block is 2.48kg?

and also, a similar question:

A pickup truck is loaded with crates of oranges. The crates have a co-efficient of static friction of 0.234 with the floor of the truck. When the truck is traveling at a speed of 46.2km/hr, in how short a distance can it come to rest without causing the crates to slide?

2. Relevant equations
Kinematic equations:
d=vit+(at^2)/2
F-Fk=ma
Fk/kN
FssN

3. The attempt at a solution
FmaxkN
Fmax=(0.625)(2.48)(9.81)
Fmax=15.2055 N

F-Fk=ma
15.2055-(9.81)(0.137)(1.57+2.48)=(2.48+1.57)a
a=2.4105 m/s^2

4.46=(0)t+((2.4105)t^2)/2
t=1.56 s

This answer is wrong, but I'm not sure what I did incorrectly.

The second question I have no idea what to do, since no masses are given, only the value of the coefficient of static friction. I tried equation F=ma and F=[μ][/s]mg, but cancelling the masses didn't make sense to me.

Last edited: Sep 26, 2015
2. Sep 26, 2015

### Staff: Mentor

I presume that the force F is applied to the lower block?

This is the maximum value of static friction between the two blocks. Good, you'll need this.

It's not clear what you are analyzing here. Both masses together? Is F the applied force? Why set it equal to the static friction?

Hint: Analyze the top block alone.

3. Sep 26, 2015

### Koru

I added the masses together because the normal force is equal to the mass of both block together, is it not? It was set equal to the force of static friction because I thought this would be the maximum force that would not move the top block. It was set equal to ma, because both block accelerate together. What is it that I have not done properly?

4. Sep 26, 2015

### Staff: Mentor

If you wanted to find the normal force with the ground, that is correct. (That normal force will equal the total weight of both blocks.)

The max value of static friction is the maximum force that can be exert on the top block (by the bottom block) without it slipping. But that's not the same as the force F.

Again: Analyze the forces acting on the top block alone.

5. Sep 26, 2015

### Koru

I'm sorry, I don't understand. Aren't any force acting on the bottom also affecting the top and vice-versa? Why is my approach incorrect?

6. Sep 27, 2015

### haruspex

The question centres on slippage between the two blocks. That depends on the forces at that interface. It doesn't "know" anything about the ground, or the interface between the lower block and the ground. The top block is accelerated by the frictional force between the two blocks. Its acceleration is a maximum when that frictional force is a maximum.

7. Sep 27, 2015

### HallsofIvy

Staff Emeritus
Which is NOT the mass of the blocks though it is proportional to it.

8. Sep 27, 2015

### Staff: Mentor

What you need to do is clearly identify what forces act on the top block. Do the same exercise for the bottom block.

Some of the forces, such as friction, you can figure out immediately; others, such as the applied force F, are initially unknown.

Realize that the only way for force F, which is applied to the bottom block, to affect the top block is through the friction force.