Finding Applied force given Mass and Kinetic and Static coefficients.

In summary, to get the newly delivered refrigerator to start moving, you will need to apply a horizontal force greater than or equal to the maximum static friction force of 370.44 N. Once the force applied exceeds this amount, the refrigerator will begin to move with an acceleration of 0.784 m/s^2. It is important to note that only one type of friction force can act at a time, either static or kinetic, and in this case, it is the maximum static friction force that must be overcome to initiate movement.
  • #1
Hieroglyphics
1
0

Homework Statement


Your family just had anew refrigerator delivered. The delivery man has left and you realize that the refrigerator is not quite in the right position, so you plan to move it several centimeters. If the refrigerator has a mass of 180 kg, the coefficient of kinetic friction between the bottom of the refrigerator of the floor is 0.13, and the static coefficient of friction between these same surfaces is 0.21, how hard do you have to push horizontally to get the refrigerator to start moving?


Homework Equations


The one's I'm thinking of are:
Fnet = ma
Fnµ = Ff

The Attempt at a Solution



So I found the Normal Force and Force of Gravity are 1764 obviously

Force of kinetic friction is 229.32, force of static friction is 370.44
and I think I can do 370.44-229.32 = Fnetx so that Fnetx = 141.12
so then I would know ma = 141.12 so a = .784.

But I'm stuck after that :(
 
Last edited:
Physics news on Phys.org
  • #2
Hieroglyphics said:

Homework Statement


Your family just had anew refrigerator delivered. The delivery man has left and you realize that the refrigerator is not quite in the right position, so you plan to move it several centimeters. If the refrigerator has a mass of 180 kg, the coefficient of kinetic friction between the bottom of the refrigerator of the floor is 0.13, and the static coefficient of friction between these same surfaces is 0.21, how hard do you have to push horizontally to get the refrigerator to start moving?


Homework Equations


The one's I'm thinking of are:
Fnet = ma
Fnµ = Ff

The Attempt at a Solution



So I found the Normal Force and Force of Gravity are 1764 obviously

Force of kinetic friction is 229.32, force of static friction is 370.44
and I think I can do 370.44-229.32 = Fnetx so that Fnetx = 141.12
so then I would know ma = 141.12 so a = .784.

But I'm stuck after that :(
You cannot have static friction and kinetic friction acting at the same time. Static friction acts only when the object is not moving with respect to the surface it is in contact with, and kinetic friction acts only when the object is moving with respect to the surface it is in contact with. Just before the refrigerator starts moving, which type friction force applies? Is there any acceleration of the refrigerator before it starts to move? What force must be applied at the instant just before it starts to move?
 

1. What is the formula for finding applied force with given mass, kinetic coefficient, and static coefficient?

The formula is F = m * (μk + μs), where F is the applied force, m is the mass, and μk and μs are the kinetic and static coefficients, respectively.

2. How do the kinetic and static coefficients affect the applied force?

The kinetic coefficient affects the applied force by determining the amount of force needed to overcome the friction between two surfaces in motion. The static coefficient, on the other hand, affects the applied force by determining the minimum amount of force needed to start an object in motion.

3. Can the applied force be negative?

Yes, the applied force can be negative if it is acting in the opposite direction of motion. This can happen if an external force is applied to slow down or stop an object.

4. What are the units of measurement for the applied force?

The units of measurement for the applied force depend on the units used for mass and the coefficients. If the mass is measured in kilograms and the coefficients are unitless, then the applied force will be in newtons (N).

5. Is there a limit to the amount of applied force that can be calculated using this formula?

There is no limit to the amount of applied force that can be calculated using this formula. However, it is important to note that the formula assumes ideal conditions and may not accurately reflect real-world scenarios with varying factors such as surface roughness and temperature.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
329
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
922
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
901
Back
Top