Object exerts an opposing force of equal magnitude

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SUMMARY

The discussion centers on Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. Participants clarify that when pushing an object, the forces involved are between the person and the object, and the object and the surface beneath it. The net force acting on the object determines its motion, specifically when the applied force exceeds static friction. Additionally, the conversation addresses calculating the coefficient of friction using the normal force and applied force, emphasizing the distinction between static and kinetic friction.

PREREQUISITES
  • Understanding of Newton's Laws of Motion, particularly Newton's Third Law.
  • Familiarity with concepts of force, friction, and acceleration.
  • Knowledge of static and kinetic friction coefficients.
  • Ability to apply basic algebra to solve physics problems involving forces.
NEXT STEPS
  • Learn how to calculate the coefficient of static friction using the formula F = μ_sN.
  • Study Newton's Second Law and its application in determining net forces.
  • Explore the differences between static and kinetic friction in practical scenarios.
  • Investigate the effects of different surfaces on frictional forces in real-world applications.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of motion and friction in mechanics.

  • #31
Ukitake Jyuushirou said:
since the weight of the sled is - 196N downwards and if there are no vertical forces, the Fn should be + 196 N but since the pull supplies a 40 N positive force, then the Fn will be 156

correct?
For your own sake:
Gain familiarity and practice with equation solving by doing this with standard mathematical formalism!
 
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  • #32
I think he will learn mathematical formalism in due time :smile:

Yes, your value for normal force is correct .
 
  • #33
arildno said:
For your own sake:
Gain familiarity and practice with equation solving by doing this with standard mathematical formalism!
wat is mathematical formalism? pardon me as i have not been doing physics in a long time and the last time i did physics was really simple and rudimentary physics. nothing beyond blind application of F = ma and the likes...

i'm pretty much trying to figure out how to study physics correctly :redface:
 
  • #34
Ukitake Jyuushirou said:
since the weight of the sled is - 196N downwards and if there are no vertical forces, the Fn should be + 196 N but since the pull supplies a 40 N positive force, then the Fn will be 156

correct?
Yes, but please do not try to do these calculations in your head! Do as arildno (and I) suggests: Write the equation for vertical equilibrium. Then solve the equation.

When you've done it formally (and correctly) a zillion times, then you can take shortcuts. :wink:
 
  • #35
Doc Al said:
Yes, but please do not try to do these calculations in your head! Do as arildno (and I) suggests: Write the equation for vertical equilibrium. Then solve the equation.

When you've done it formally (and correctly) a zillion times, then you can take shortcuts. :wink:
ok got it :smile::smile:

w=mg
196=20x9.8

Fn = 196 (no vertical force)

196-40 = 156 (force of 40N in +y-component)
Fn = 156
 
  • #36
now i have to solve for the coefficient of kinetic friction. kinetic friction force is supposed to be proportional to the Fn. but from wat is given, how do i work out the friction force?
 
  • #37
What is the constant of proportionality in the equation relating kinetic frictional force and normal force ?
What are the conditions for horizontal equilibrium ?
As Doc Al suggested, write down the math step by step .
 
  • #38
frictional force is related to the Fn.

Fn = 156, Fx = 69.2. coefficient of friction is 69.2 divided by 156 = 0.44

am i correct?
 
  • #39
Ukitake Jyuushirou said:
w=mg
196=20x9.8

Fn = 196 (no vertical force)

196-40 = 156 (force of 40N in +y-component)
Fn = 156
Good. But just for the record, here's how I would do it:
(Sum of the forces in the vertical direction) = +Fn -mg +80sin(30) = 0

That's the equation for vertical equilibrium. Solving it (trivial, yes) gives:
Fn = mg - 80sin(30)

Now do the same for horizontal forces.
 
  • #40
Doc Al said:
Good. But just for the record, here's how I would do it:
(Sum of the forces in the vertical direction) = +Fn -mg +80sin(30) = 0

That's the equation for vertical equilibrium. Solving it (trivial, yes) gives:
Fn = mg - 80sin(30)

Now do the same for horizontal forces.
using ur formula


Fx = 80cos30 = 69.2
 
  • #41
OK. Again, the way I would do it is:
(Sum of the forces in the horizontal direction) = -Ff +80cos(30) = 0

Solving gives: Ff = 80cos(30)

(The reason for sticking to a systematic approach is: (1) It drives home the key physics principles & (2) It allows you to solve much harder problems.)

Now you can apply the other relationship you know about kinetic friction to solve for \mu_k systematically.
 
  • #42
Doc Al said:
OK. Again, the way I would do it is:
(Sum of the forces in the horizontal direction) = -Ff +80cos(30) = 0

Solving gives: Ff = 80cos(30)

(The reason for sticking to a systematic approach is: (1) It drives home the key physics principles & (2) It allows you to solve much harder problems.)

Now you can apply the other relationship you know about kinetic friction to solve for \mu_k systematically.
yes ur right. :smile: i'd try to use this approach for my other qns

frictional force is related to the Fn.

Fn = 156, Fx = 69.2. coefficient of friction is 69.2 divided by 156 = 0.44

am i correct?
 
  • #43
Looks good. I'd write it this way. The key relationship for kinetic friction is:
F_f = \mu_k F_n

thus:
\mu_k = F_f /F_n

then just plug in the numbers
 
  • #44
whew...one question down...many many more to go... thanks for everyone's help and advice :smile:
 
  • #45
I think we beat that one to death! :smile:
 
  • #46
We sure did :smile:
 

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