Understanding the Direction of Friction Forces in Equilibrium

In summary, the direction of friction on the wire is upwards, and the direction of friction on the ring is downwards. This is determined by considering the equilibrium of the system and the forces acting on the ring and the wire. The force of friction is necessary to keep the ring in place and counteract the vertical forces acting on it. The force of friction is also affected by the angle of the tension force, with mgcos(60°) being less than mg, and 3mgcos(30°) being greater than 2mg. This explains the difference in directions for the two problems.
  • #1
synkk
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ixFM5.png


[strike]In this question, the direction of friction (which I assumed correctly) is acting downwards on the wire,[/strike] however:

bKtet.png


The friction is acting upwards on the wire, could someone walk me through how I would figure out the direction of the friction force?

I understand friction is against motion, but I don't see the motion.
 
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  • #2
synkk said:
ixFM5.png


In this question, the direction of friction (which I assumed correctly is acting downwards on the wire, however:

bKtet.png


The friction is acting upwards on the wire, could someone walk me through how I would figure out the direction of the friction force?

I understand friction is against motion, but I don't see the motion.

In the first example, a free body diagram for the particle (the black circle) along with the statement that the system is in equilibrium, tells you that T = mg, where T is the tension in the rope. Imposing the requirement of vertical equilibrium for the ring, you see that the downward force on it is mg, but the upward force on it due to the tension in the string is only mgcos(60°) < mg. So, the ring wants to slide downward. In order for the ring to be static, there must be an upward frictional force that keeps it suspended by making up the balance of the vertical force.

So, the force of friction on the ring due to the wire is upwards. Therefore, by Newton's 3rd Law, the force of friction on the wire due to the ring is downwards, just as you stated in your original post.

In contrast, for the second problem, the vertical component of the tension in the rope is 3mgcos(30°) > 2mg. So the force pulling upward on the ring is greater than its weight. It wants to slide upward, and it would, were it not for a downward frictional force with the wire that keeps it in place.

So, the force of friction on the ring due to the wire is downwards. Therefore, by Newton's 3rd Law, the force of friction on the wire due to the ring is upwards, just as you stated in your original post.
 
  • #3
cepheid said:
In the first example, a free body diagram for the particle (the black circle) along with the statement that the system is in equilibrium, tells you that T = mg, where T is the tension in the rope. Imposing the requirement of vertical equilibrium for the ring, you see that the downward force on it is mg, but the upward force on it due to the tension in the string is only mgcos(60°) < mg. So, the ring wants to slide downward. In order for the ring to be static, there must be an upward frictional force that keeps it suspended by making up the balance of the vertical force.

So, the force of friction on the ring due to the wire is upwards. Therefore, by Newton's 3rd Law, the force of friction on the wire due to the ring is downwards, just as you stated in your original post.

In contrast, for the second problem, the vertical component of the tension in the rope is 3mgcos(30°) > 2mg. So the force pulling upward on the ring is greater than its weight. It wants to slide upward, and it would, were it not for a downward frictional force with the wire that keeps it in place.

So, the force of friction on the ring due to the wire is downwards. Therefore, by Newton's 3rd Law, the force of friction on the wire due to the ring is upwards, just as you stated in your original post.

I'm sorry but the bold part, how did you get it less than mg? Also, I've looked at the solutions and for the first 1, they have the friction upwards, and downwards for the second one, so it seems I was wrong.
 
  • #4
synkk said:
I'm sorry but the bold part, how did you get it less than mg?

For purely geometrical reasons, mgcos60 is less than mg. cepheid stated this in order to then say that the force of friction must be what's holding everything in equilibrium.
 
  • #5
synkk said:
I'm sorry but the bold part, how did you get it less than mg?

Yeah, what gnulinger said. cos(any angle) ≤ 1.
synkk said:
Also, I've looked at the solutions and for the first 1, they have the friction upwards, and downwards for the second one, so it seems I was wrong.

They are talking about the friction acting ON THE RING. You were talking about the friction acting ON THE WIRE. These are in opposite directions according to Newton's 3rd law. I already explained that above. I gave the directions for both of the forces in the action-reaction pair in both examples.
 
  • #6
gnulinger said:
For purely geometrical reasons, mgcos60 is less than mg. cepheid stated this in order to then say that the force of friction must be what's holding everything in equilibrium.

cepheid said:
Yeah, what gnulinger said. cos(any angle) ≤ 1.




They are talking about the friction acting ON THE RING. You were talking about the friction acting ON THE WIRE. These are in opposite directions according to Newton's 3rd law. I already explained that above. I gave the directions for both of the forces in the action-reaction pair in both examples.
thanks both, get it now.
 

1. What is the direction of friction?

The direction of friction is always opposite to the direction of motion or the force applied.

2. How does the direction of friction affect an object's movement?

The direction of friction can either slow down or stop an object's movement, depending on the direction and magnitude of the applied force.

3. What factors determine the direction of friction?

The direction of friction is determined by the surface materials, the angle of the surface, and the type of motion (sliding, rolling, or static).

4. Can the direction of friction change?

Yes, the direction of friction can change depending on the changing conditions such as the angle of the surface or the applied force.

5. How does the direction of friction affect the coefficient of friction?

The direction of friction has no direct effect on the coefficient of friction, which is a measure of the force needed to keep an object in motion. However, the coefficient of friction can change if the direction or magnitude of the applied force changes.

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