Solve Newtons 2nd Law for Friction Coefficient: 4kg School Bag, 20N Force

In summary, a student wonders about the coefficient of friction while dragging a 4kg school bag at a constant speed with a force of 20N at an angle of 60 degrees above the horizontal. After finding the vertical and horizontal components of the force, the student calculates the coefficient of friction to be approximately 0.46.
  • #1
purplelgnd18
5
0

Homework Statement


While walking down the hallway, a student drags his 4kg school bag. The force on the bag is 20N [60 degrees above the horizontal]. If the bag moves at a constant speed, what is the coefficient of friction?


Homework Equations


Fn = Fg-Fasintheta

Ff = u(Fn).


The Attempt at a Solution


Well, I found Fg to be aprrox. 39.2N, then I found Fn to be approx. 21.9N. Now I'm stumped. Any help would be appreciated.
 
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  • #2
Welcome to PF, Purplelgnd!
Looks like you have the vertical part all figured out correctly. You just need the horizontal applied force and then you can write F = ma for the horizontal. There will, of course, be two horizontal forces to add or subtract and one of them will have the μ you are looking for in its expression.
 
  • #3
Delphi51 said:
Welcome to PF, Purplelgnd!
Looks like you have the vertical part all figured out correctly. You just need the horizontal applied force and then you can write F = ma for the horizontal. There will, of course, be two horizontal forces to add or subtract and one of them will have the μ you are looking for in its expression.

Thanks for the welcome! Purplelgnd is actually a name I made up for a Minnesota Vikings forum that I post on (not a proud day today to be a fan I admit).

OK but sadly, I'm still stumped. I'm actually about to break down into tears right now, LOL.
 
  • #4
Ok so I plugged in a few numbers and tried this.

Ff = -Facos60
Ff = -(20N)cos60
Ff = 10N

Then:

u = Ff/Fn
u= 10N / 21.9 N
u = ~0.49

Is this correct?
 
  • #5
Start by finding the horizontal part of the 20 N force at angle 60 degrees.

Pardon my offering unsolicited advice, but it really, really helps you, me and your poorly paid marker if you get in the habit of writing things out nicely. For the part you already did finding Fn = 21.9, you should have a diagram showing the 20 N at an angle of 60 degrees. Then show it replaced by a vertical plus a horizontal vector.
bag.jpg

Next step, find Fx.
 
  • #6
Okay, you beat me to it. But check that 10/21.9 calc again - I don't get .49.
 
  • #7
I think I was reading the numbers off wrong.

~0.46?
 
  • #8
That's it!
Good luck on the next one.
 
  • #9
haha, Thank you so much!

My dad is a Physics Prof. over at the University of Waterloo (in Ontario), and he didn't have time to help me with this homework. LOL, oh well.

Thanks again!
 

Related to Solve Newtons 2nd Law for Friction Coefficient: 4kg School Bag, 20N Force

1. What is Newton's 2nd Law?

Newton's 2nd Law of Motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. In other words, the greater the force applied to an object, the greater its acceleration will be, and the more massive the object, the less its acceleration will be.

2. How do you solve for the friction coefficient?

To solve for the friction coefficient in this scenario, we can use the formula μ = F/N, where μ is the coefficient of friction, F is the force of friction, and N is the normal force. In this case, the force of friction is equal to the force applied (20N), and the normal force is equal to the weight of the object (4kg x 9.8m/s² = 39.2N). Plugging these values into the formula, we get μ = 20N/39.2N = 0.51.

3. Why is the mass of the object important in this calculation?

The mass of the object is important because it affects the acceleration of the object. As stated in Newton's 2nd Law, the acceleration of an object is inversely proportional to its mass. In this scenario, the heavier the object, the less it will accelerate, and the lighter the object, the more it will accelerate. This is why we need to know the mass of the object in order to accurately calculate the friction coefficient.

4. How does the force of the school bag affect the friction coefficient?

The force of the school bag directly affects the friction coefficient because it is one of the variables in the formula μ = F/N. As the force of the school bag increases, the friction coefficient will also increase, assuming the normal force remains constant. This means that the heavier the bag is, the more force it will exert on the surface, resulting in a higher coefficient of friction.

5. Can the friction coefficient change in different situations?

Yes, the friction coefficient can change in different situations. It is dependent on various factors such as the type of surface, the smoothness of the surface, and the weight of the object. For example, the friction coefficient will be different for a heavy object on a rough surface compared to a light object on a smooth surface. This is why it is important to consider all the variables and conditions when calculating the friction coefficient.

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