Distance to Come to Rest and Force of Friction

In summary: So, in a sense, the top block "knows" about the applied force only through the friction force.In summary, the problem involves two blocks with different masses and coefficients of friction. The question is asking for the minimum amount of time it would take for both blocks to travel a distance of 4.46m without the top block sliding on the lower block. To solve this, we need to analyze the forces acting on the top block alone, including the applied force F and the friction force between the two blocks. The maximum value of static friction between the two blocks is 15.2055 N. To find the acceleration of the top block, we can use the equation F-Fk=ma and solve for a.
  • #1
Koru
3
0

Homework Statement


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?

Homework Equations


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

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.
 
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  • #2
Koru said:
Force F causes both blocks to cross a distance of 4.46m, starting from rest.
I presume that the force F is applied to the lower block?

Koru said:
Fmax=μ/kN
Fmax=(0.625)(2.48)(9.81)
Fmax=15.2055 N
This is the maximum value of static friction between the two blocks. Good, you'll need this.

Koru said:
F-Fk=ma
15.2055-(9.81)(0.137)(1.57+2.48)=(2.48+1.57)a
a=2.4105 m/s^2
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
Doc Al said:
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.

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
Koru said:
I added the masses together because the normal force is equal to the mass of both block together, is it not?
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.)

Koru said:
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.
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
Doc Al said:
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.

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
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
Doc Al said:
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.)
Which is NOT the mass of the blocks though it is proportional to it.
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.
 
  • #8
Koru said:
Aren't any force acting on the bottom also affecting the top and vice-versa?
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.
 

Related to Distance to Come to Rest and Force of Friction

1. What is distance to come to rest?

Distance to come to rest refers to the distance an object travels before it comes to a complete stop. It is affected by the object's initial velocity, the force of friction, and the mass of the object.

2. How is distance to come to rest calculated?

The formula for distance to come to rest is (initial velocity)^2 / (2 * coefficient of friction * acceleration due to gravity). This assumes that the only force acting on the object is friction.

3. What factors affect the force of friction?

The force of friction is affected by the type of surface the object is moving on, the weight of the object, and the coefficient of friction between the object and the surface.

4. How does the force of friction impact an object's distance to come to rest?

The force of friction acts in the opposite direction of an object's motion, slowing it down. This means that a higher force of friction will result in a shorter distance to come to rest.

5. Can the distance to come to rest be greater than the distance traveled?

Yes, the distance to come to rest can be greater than the distance traveled if there are other forces acting on the object, such as air resistance or an inclined plane. In these cases, the object may travel a longer distance before coming to a complete stop.

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