Exploring the Mysteries of Friction: Newton's 3rd Law & Tug-of-War

In summary, the coefficient of friction does not have any units because it is a constant derived from an ODE. In a tug-of-war match, Newton's 3rd law can be used to win by loosening grip, weighing less, or being weaker. To determine if a rope will break, the tension in the rope must be equal to or less than its breaking point. The formula for friction force is F=muN and the coefficient of friction is a unitless constant. When solving for the weight of a fish using Newton's 2nd law, the units must be consistent.
  • #1
Fusilli_Jerry89
159
0
1) Why doesn't the coefficient of friction have any units?

2) A rop can withstand 750 Newtons. If 2 people in opposite directions pull with a force of 500 Newtons will the rope break?

3) In a tug-of-war Newtons 3rd law states that any action causes an equal yet opposite reaction. Explain how a tug-of-war match can be won using this law. I put because one side may loosen their grip, or weigh less and therefore have less friection on their side, or may be weaker so thery cannot possible apply the force the other side can without sliding.

4) A fisherman uses a line that has a test value of 50 pounds. If he pulls the fish upwards at 2m/s/s how much does the fish weigh at least? I put 25 lbs.
 
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  • #2
Fusilli_Jerry89 said:
1) Why doesn't the coefficient of friction have any units?

2) A rop can withstand 750 Newtons. If 2 people in opposite directions pull with a force of 500 Newtons will the rope break? Draw a diagram of the rope. Then apply Newton's laws using a Free body diagram of the system, and a FBD cut through the rope that includes one of the pulling forces on the left or right part of the rope.

3) In a tug-of-war Newtons 3rd law states that any action causes an equal yet opposite reaction. Explain how a tug-of-war match can be won using this law. I put because one side may loosen their grip, or weigh less and therefore have less friection on their side, or may be weaker so thery cannot possible apply the force the other side can without sliding.

4) A fisherman uses a line that has a test value of 50 pounds. If he pulls the fish upwards at 2m/s/s how much does the fish weigh at least? I put 25 lbs. I think you mean how much does the fish weigh at most. Again, do a FBD and apply Newton's 2nd law.
See comments above in red.
 
  • #3
so the rope wouldn't break, i donno how but I got about 40 lbs. for the fish, but why duznt the coefficient have units?
 
  • #4
Fusilli_Jerry89 said:
so the rope wouldn't break, i donno how but I got about 40 lbs. for the fish, but why duznt the coefficient have units?
Why doesn't the rope break?

That fish problem has messed up units, the line strength is given as 50 pounds, but you give the acceleration in meters/s/s? Did you mean a = 2ft./s/s? If not, best to convert pounds to Newtons before solving. At any rate, use Newton's 2nd law and set the rope strength equal to the line tension to solve for the max weight. Show work.

What's the formula for the friction force?
 
  • #5
Fusilli_Jerry89 said:
so the rope wouldn't break, i donno how but I got about 40 lbs. for the fish, but why duznt the coefficient have units?

Think about the equation of friction: F = mu N. F and N have units of Newtons. Solve for mu and ask yourself what the units would be.

Sometimes, I think it's easier to deal with something more familiar. So, if I drive 4 miles on thursday, and 8 miles on Friday, I drove twice as far. What are the units of 'twice'?

Dorothy
 
  • #6
so it's just a ratio right?
 
  • #7
PhanthomJay said:
Why doesn't the rope break?

That fish problem has messed up units, the line strength is given as 50 pounds, but you give the acceleration in meters/s/s? Did you mean a = 2ft./s/s? If not, best to convert pounds to Newtons before solving. At any rate, use Newton's 2nd law and set the rope strength equal to the line tension to solve for the max weight. Show work.

What's the formula for the friction force?

well if the other end of the rope were attached to a wall, and you pulled at 500 N, and its breaking point was 750, would it break? or if would u only have to pull at half of 750 N in order to break it?. Isn't it the same concept since the wall exerts 500 N back, so does the other team. Is is the magnitude of the smallest direction of force that determines the break?
 
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  • #8
Also about the units in the formula. The coeffiecient of friction is a constant that is derived from an ODE (which you may or may not know). The result from the ODE is that the constant (the coefficient of friction) is unitless.
 
  • #9
PhanthomJay said:
Why doesn't the rope break?

That fish problem has messed up units, the line strength is given as 50 pounds, but you give the acceleration in meters/s/s? Did you mean a = 2ft./s/s? If not, best to convert pounds to Newtons before solving. At any rate, use Newton's 2nd law and set the rope strength equal to the line tension to solve for the max weight. Show work.

What's the formula for the friction force?
I came up with F-222.754=2m, but now what do I do? Do i divide 222.754 by 9.8 and use that as the mass? If so the answer would be 27 kg. And yes the question says 2m/s/s but also 50 lbs.
 
  • #10
Fusilli_Jerry89 said:
well if the other end of the rope were attached to a wall, and you pulled at 500 N, and its breaking point was 750, would it break? or if would u only have to pull at half of 750 N in order to break it?. Isn't it the same concept since the wall exerts 500 N back, so does the other team. Is is the magnitude of the smallest direction of force that determines the break?
Whether it's 2 people pulling on the rope with a force of 500 each, or one pulling on it at 500 while it is attached to a wall at the other end, yes, correct, tension in rope is the same. What is that tension? If it's less than 750, the rope will hold.
 
  • #11
Fusilli_Jerry89 said:
I came up with F-222.754=2m, but now what do I do? Do i divide 222.754 by 9.8 and use that as the mass? If so the answer would be 27 kg. And yes the question says 2m/s/s but also 50 lbs.
it's 222 - F = 2m, where F is the weight, mg, = to 9.8m. Now solve for m.
 

Related to Exploring the Mysteries of Friction: Newton's 3rd Law & Tug-of-War

1. What is friction?

Friction is a force that resists the relative motion between two objects in contact. It is caused by the irregularities in the surfaces of the objects and the interlocking of their microscopic features.

2. How does Newton's 3rd law relate to friction?

Newton's 3rd law states that for every action, there is an equal and opposite reaction. In the case of friction, when two objects are in contact, the force of friction acting on one object is equal and opposite to the force of friction acting on the other object.

3. What factors affect the amount of friction between two objects?

The amount of friction between two objects depends on the types of surfaces in contact, the force pressing the surfaces together, and the presence of any lubricants or contaminants on the surfaces.

4. How does friction impact everyday activities?

Friction plays a significant role in our daily lives. It allows us to walk, drive, and hold objects, among other things. However, it can also cause wear and tear on surfaces and objects, which can be both beneficial and harmful.

5. How can we decrease friction between two surfaces?

One way to decrease friction between two surfaces is by using a lubricant, such as oil or grease, to reduce the contact between the surfaces. Additionally, polishing or smoothing the surfaces can also decrease friction. Another method is by using wheels or rollers, which reduce the amount of surface area in contact and therefore decrease the force of friction.

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