Magnitude of static friction force

In summary, the magnitude of the static friction force between the horse's hooves and the soil is 1860 N.
  • #1
pinkpistachios
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Homework Statement


A 500 kg ox runs and speeds up by 3 m/s2, but is held back by a 120 kg plow that is attached to its back. Assuming the coefficient of static friction between the horse's hooves and the soil is 0.73, what is the magnitude of the static friction force the soil exerts on the horse's hooves (not maximum static friction) which explains the horse's acceleration?

Homework Equations


Fsfr ≤ μFN
FN=mg

The Attempt at a Solution



FN = 500(9.8) = 4900 N

Fsfr ≤ 0.73(4900)
max static friction = 3600 N

^I got up until here. I'm confused about finding the magnitude of the static friction force...do I have to find the force the plow exerts on the horse and subtract that from 3600 N? If so, how would I find that force?
 
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  • #2
You could find the force between the ox and plow, but it is not necessary to solve this problem.
You have a typo...the acceleration is 3 m/s^2.
You are missing a very important relevant equation. Which one?
 
  • #3
I have acceleration and mass, so would I use Newton's second law a = ∑F/m?
 
  • #4
3 = (- Fplow on ox + Fsfr )/500
1500 = (- Fplow on ox + Fsfr)
1500N would be the net horizontal force right? Would that be the magnitude of Fsfr ?
 
  • #5
pinkpistachios said:
3 = (- Fplow on ox + Fsfr )/500
1500 = (- Fplow on ox + Fsfr)
1500N would be the net horizontal force right? Would that be the magnitude of Fsfr ?
No.
You have calculated the net horizontal force acting on the ox correctly, but the net horizontal force acting on the ox includes the sum of both the static friction force and the force between the ox and plow. So you would have to calculate the force between the ox and plow to find the friction force. But instead, apply Newtons law to the entire ox-plow system. Then you don't need that force between ox and plow because it is internal to the system.
 
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  • #6
If I combine the masses so it's one system and then apply N2L:
3 = F / 620
so 1860 = Fsfr ?
 
  • #7
pinkpistachios said:
If I combine the masses so it's one system and then apply N2L:
3 = F / 620
so 1860 = Fsfr ?
Yes. But don't forget units. The static friction force is 1860 N. (If you wish, for extra credit, you might want to calculate the force between that ox and plow).
 
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  • #8
I got it, thanks :)
 

Related to Magnitude of static friction force

What is the magnitude of static friction force?

The magnitude of static friction force is the maximum amount of force that must be overcome to move an object resting on a surface. It is dependent on factors such as the mass and surface roughness of the object, as well as the coefficient of static friction between the two surfaces.

How is the magnitude of static friction force calculated?

The magnitude of static friction force can be calculated using the equation Fs = μs * N, where Fs is the static friction force, μs is the coefficient of static friction, and N is the normal force exerted on the object by the surface.

What factors affect the magnitude of static friction force?

The magnitude of static friction force is affected by the mass and surface roughness of the object, as well as the coefficient of static friction between the two surfaces. It is also dependent on the normal force exerted on the object by the surface.

How does the magnitude of static friction force differ from kinetic friction force?

The magnitude of static friction force is the maximum amount of force that must be overcome to move an object at rest, while kinetic friction force is the force that opposes the motion of an object that is already in motion. Additionally, the coefficient of kinetic friction is typically lower than the coefficient of static friction.

Why is understanding the magnitude of static friction force important?

Understanding the magnitude of static friction force is important in many real-world applications, such as designing brakes for vehicles, calculating the force needed to push or pull an object, and ensuring the safety of structures and structures. It also allows us to predict the behavior of objects on different surfaces and make informed decisions in engineering and physics.

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