Magnitude of static friction force

AI Thread Summary
The discussion focuses on calculating the magnitude of the static friction force exerted on a 500 kg ox accelerating at 3 m/s² while pulling a 120 kg plow. The relevant equations include the normal force and static friction force, with the coefficient of static friction given as 0.73. The participants clarify that the net horizontal force acting on the ox is the sum of the static friction force and the force between the ox and plow. By applying Newton's second law to the entire ox-plow system, they determine that the static friction force is 1860 N. The conversation emphasizes the importance of using the correct approach to simplify the problem.
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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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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?
 
I have acceleration and mass, so would I use Newton's second law a = ∑F/m?
 
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 ?
 
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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If I combine the masses so it's one system and then apply N2L:
3 = F / 620
so 1860 = Fsfr ?
 
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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I got it, thanks :)
 
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