# How Do You Calculate Net Force and Coefficient of Friction in Physics Problems?

• max1020
In summary: No, net force and normal force are not the same. Net force is the total force acting on an object, while normal force is the force perpendicular to the surface that an object is resting on.For problem 3a, the magnitude of the frictional force can be calculated by multiplying the coefficient of friction (0.37) by the normal force (FN). So the equation would be Ff = µ * FN. In this case, Ff = 0.37 * 28N = 10.36N.
max1020
these are technically not homework, i just need more help for my test tomorrow!
I HAVE NO IDEA HOW TO DO THIS! can someone walk me through these and show me the steps! thank you :)

1.A sled is being pulled to the right with a force of 28 N. A frictional force is working opposite the applied force with a magnitude of 17 N. The mass of the sled is 4.5 kg. (a) What is the net force in the x direction? 28 – 17 = 11 N (b) What is the coefficient of friction between the sled’s blades and the surface? m = 0.385

2. A football player with a mass of 86 kg, dives into the end zone and is slowed by the grass after he hits the ground by a force of friction of 580 N. What is µ between the player and the grass? m = 0.69

3. A person pushes horizontally with a force of 350 N on a 68 kg chair to move it across a level floor. The coefficient of friction is 0.37. (a) What is the magnitude of the frictional force? 247 N (b) What is the acceleration of the chair? Fnet = 350 – 247 = 103 N, acceleration = 1.51 m/s2

max1020 said:
I HAVE NO IDEA HOW TO DO THIS!
Sorry but you need to at least show an attempt.
You need to try to understand what it is about the problem that you don't understand.

This is copied from the thread at the top of this (Introductory Physics Homework) forum:
Guidelines for Students and Helpers said:
Show us that you've thought about the problem.
The forum rules require that you show an attempt at solving the problem on your own. Obviously, one reason we want to see your work is because we prefer to help those who are genuinely trying and interested in learning. What's more important is that we need to see what you've tried so we know how to help you. For your attempt, you can offer a partial solution to the problem, but you don't always have to. What we're really interested in is seeing what you're thinking so we can identify and clear up any misconceptions or points of confusion.

Here are some common mistakes to avoid:

Don't simply say "I have no clue," "I have no idea where to start," or "I'm completely lost." These don't qualify as attempts. Instead, it suggests you haven't put much effort into reading and understanding your textbook and lecture notes, going over similar examples, etc. The helpers aren't here to answer your questions so you don't have to read your book.
Don't simply say "I tried for hours and didn't get anywhere." This is really no better than saying "I have no clue." It tells us absolutely nothing about where you're getting stuck. If you tried for hours, you must have had some thoughts about the problem. What were they? Show us what you tried, explain why you think it was wrong, and so on. Better yet, identify what's confusing you and ask specific questions to help you figure things out.
Don't just give a vague or general description of what you tried. If the problem lies in the execution, as it often does, we can't help you find mistakes without seeing your work in detail. Even if you provide your final result, it usually does little to help us figure out where your attempt went awry.

max1020 said:
these are technically not homework, i just need more help for my test tomorrow!
I HAVE NO IDEA HOW TO DO THIS! can someone walk me through these and show me the steps! thank you :)

1.A sled is being pulled to the right with a force of 28 N. A frictional force is working opposite the applied force with a magnitude of 17 N. The mass of the sled is 4.5 kg. (a) What is the net force in the x direction? 28 – 17 = 11 N (b) What is the coefficient of friction between the sled’s blades and the surface? m = 0.385

2. A football player with a mass of 86 kg, dives into the end zone and is slowed by the grass after he hits the ground by a force of friction of 580 N. What is µ between the player and the grass? m = 0.69

3. A person pushes horizontally with a force of 350 N on a 68 kg chair to move it across a level floor. The coefficient of friction is 0.37. (a) What is the magnitude of the frictional force? 247 N (b) What is the acceleration of the chair? Fnet = 350 – 247 = 103 N, acceleration = 1.51 m/s2
Sorry I did my work on appear and was too lazy to write them all online. For the coefficient question do I do 11N/28N

max1020 said:
For the coefficient question do I do 11N/28N

No that's not right. What is coefficient of friction? Doesn't the force of friction depend on how hard two objects are pressing against each other?

Nathanael said:
No that's not right. What is coefficient of friction? Doesn't the force of friction depend on how hard two objects are pressing against each other?

So is Ff 11N and FN 28N? Can you tell me how to identify the Ff and FN if I did it wrong

max1020 said:
So is Ff 11N and FN 28N? Can you tell me how to identify the Ff and FN if I did it wrong

$F_{friction}$ is said to be 17N in the problem
("A frictional force is working ... with a magnitude of 17N")

(By "FN" you mean the normal force, right?)

$F_{normal}$ is the force that the ground is pushing up on the sled
(which is equal to the force with which the sled is pushing on the ground)

P.S.

so it would be 17/11
which is 1.54

is net force the same as normal force

and for number 3a, the answer was online but i didn't get how to do it. i used Ff=(μ)(FN) and i got Ff=(.37)(350N) BUT my answer was 129.5 when the answer should be 247N

max1020 said:
is net force the same as normal force

Nope. Right now, gravity is applying a force on you, but you're not accelerating. The reason is because there is another force acting on you (equal in size and opposite in direction) from the ground (or from a chair or something) and it is called the normal force.

To find the normal force, you note that the sled is not vertically accelerating, so there must be a net force of zero in the vertical direction.
(So $F_{gravity}+F_{normal}=0$)

so i have to do fnormal=fnet-fgravity which is 9.8N

max1020 said:
and for number 3a, the answer was online but i didn't get how to do it. i used Ff=(μ)(FN) and i got Ff=(.37)(350N) BUT my answer was 129.5 when the answer should be 247N

Well that's because you used the wrong value for the normal force.

Kinetic friction does not depend on how hard you're pulling on the object, it depends on how hard the two object and the ground are pressed together (the normal force).

max1020 said:
so i have to do fnormal=fnet-fgravity which is 9.8N

So you're saying Fgravity is (negative) 9.8N?

But isn't Fgravity=mg?

The mass of the sled is not 1kg

i got it! can u explain 3.a please? THANKS

max1020 said:
i got it! can u explain 3.a please? THANKS

You had the right idea you just used the wrong normal force:
max1020 said:
and for number 3a, ... i used Ff=(μ)(FN) and i got Ff=(.37)(350N)

You're given the mass of the chair so what is the normal force?

1 person

## What is frictional force?

Frictional force is the resistance force that occurs when two surfaces come in contact and slide against each other. It is caused by the microscopic roughness of the surfaces and the interlocking of the irregularities.

## How is frictional force calculated?

Frictional force can be calculated using the equation F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force between the two surfaces.

## What factors affect frictional force?

The factors that affect frictional force include the type of surfaces in contact, the roughness of the surfaces, the force pressing the surfaces together, and the presence of any lubricants or contaminants.

## Why is understanding frictional force important?

Understanding frictional force is important because it plays a crucial role in many everyday activities, such as walking, driving, and using tools. It also has important applications in engineering, physics, and other scientific fields.

## How can frictional force be reduced?

Frictional force can be reduced by using lubricants, polishing surfaces, or changing the type of surfaces in contact. It can also be reduced by minimizing the force pressing the surfaces together, such as using wheels or ball bearings.

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