# Frictional force word problem test tomorrow!

1. Aug 6, 2014

### 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

2. Aug 6, 2014

### Nathanael

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:

3. Aug 6, 2014

### max1020

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

4. Aug 6, 2014

### Nathanael

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?

5. Aug 6, 2014

### max1020

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

6. Aug 6, 2014

### Nathanael

$F_{friction}$ 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.

7. Aug 6, 2014

### max1020

so it would be 17/11
which is 1.54

8. Aug 6, 2014

### max1020

is net force the same as normal force

9. Aug 6, 2014

### max1020

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

10. Aug 6, 2014

### Nathanael

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 $F_{gravity}+F_{normal}=0$)

11. Aug 6, 2014

### max1020

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

12. Aug 6, 2014

### Nathanael

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).

13. Aug 6, 2014

### Nathanael

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

But isn't Fgravity=mg?

The mass of the sled is not 1kg

14. Aug 6, 2014

### max1020

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

15. Aug 6, 2014

### Nathanael

You had the right idea you just used the wrong normal force:
You're given the mass of the chair so what is the normal force?

16. Aug 6, 2014