Question

what is the force the runner is pushing the defensive back with? Runners mass=100kg, Db mass=90 kg, his applied force is 212 newtons and coefficient of friction=.2. There is constant velocity. Also, are the forward/backward/stopped/what is frictional force of each/ normal force/weight.
Thank you

Answer 1

Might need a picture. Why are the forces not equal and opposite or are they?

Answer 2

Uh they arent on a ramp

Answer 3

so normal force =wieght right?

Answer 4

If they are on a ramp, then its angle @ makes a difference.
If it is flat on the ground, no slope, then the normal force is the weight.

If it is at angle @ then the normal force from the weight is W cos@..

Answer 5

I mean im assuming its flat because its on a football field

Answer 6

Fine, yes horizontal, flat. I was not sure an angle was involved, for some reason. Since no acceleration is taking place the frictional force equals the force they each exert, so it is the coefficient of friction times the normal force for each player. The heavier player can apply more force without slipping.

Answer 7

So...whats the answer to the question?Im kind of new to physics..

Answer 8

Can anyone answer my question?

Answer 9

The goal here is not to give answers, but to help people get answers.

If they are pushing on each other with the same force and the heavier man slip less because the friction is proportional to his weight, then which way will they slide if they slide? They will slide if the applied force between them is greater than the frictional force the less man can apply. This is a hard problem for someone new to physics.

Advice: make a picture, draw arrows for force magnitudes and directions, list what you know and what you want to know.

Answer 10

I know, I just joined this mechanical engineering program and they skipped over alot of material, everyone takes physics in it except me, it take biology, so this is kinda challenging for me.

Answer 11

ok so
since the weight of the runner is 100 kg then thats 1000 newtons normal force/weight?

Answer 12

and the coefficient of friction is .2 so 1000n times that is 200n so thats the friction

Answer 13

and the force of the DB is 212 newton so would i subtract 200 from 212?

Answer 14

this is what i mainly am confused about

Answer 15

Anyone there??

Answer 16

Who does "his" refer to? Runner or back? Anyway, yes, calculate the friction forces, and compare against the 212N between the two men. Use 9.8 N/kg rather than 10N/kg for g.

Answer 17

why 9.8?, just wondering and the applied force is refering to the runner i assume because the other Db's force is already given

Answer 18

?

Answer 19

First calculate the Normal force for the Db.
F = ma Where m is the mass, 90 kg, and a is the acceleration due to gravity, 9.8 m/s^2
That will be the Normal force in Newtons.
Since the question tell us the Db is moving, we only have to look at the kinetic frictional force, Fk
Fk will be = the coefficient of friction x the normal force in this situation:
\[F_{k} = \mu _{k} \times F_{n}\]

Answer 20

The normal force is 900 newtons

Answer 21

or if you are using 9.8 its 882 newtons

Answer 22

So, take the normal force you calculated above (90 kg x 9.8 m/s^2) by the coefficient of friction, 0.2, and you have the kinetic force of friction.
It is the force pair to the force the runner is applying. (It is in the opposite direction of the 212 N the runner is applying.)

Answer 23

ok so thats176.4 newtons

Answer 24

so does this mean the (applied) force is 35.6?

Answer 25

The only thing I don't like about this one is that we are treating the Db as if he were a 'block' or other object. (Not even considering that he may be trying to apply force himself and only worrying about his mass, the normal force, and the friction.) But, there is nothing in the question that provides information to do anything other than this.)

Answer 26

yes thats why i was confused

Answer 27

Applied force is 212 N. I would probably think of the 35.6 N as a net force.

Answer 28

I mean the runner applied 35.6 n?

Answer 29

He applied 212 N, that is given. But, since the force of friction is an opposing force, the net (sum) of the forces is 35.6 N

Answer 30

the sum is 35.6? I subtracted 176.6 from 212 to get that

Answer 31

Just like if you push on a wall with 30 N of force and the wall pushes back with an equal and opposite force, the net (sum ) is 0 N. So neither you nor the wall move. But, that does not mean your applied force is any less. (The applied force is still 30 N. It just 'adds up' to 0 N after you take the opposing force into account.)

Answer 32

The force of friction does not reduce the applied force. It just tells us what the force opposing the applied force will be.

Answer 33

So how doI find the force of the runner on the db?

Answer 34

Sorry im very new to physics

Answer 35

started a day ago

Answer 36

It is given in the question. He applies a force of 212 N

Answer 37

i am talking about the force of the runner on the db?

Answer 38

Not db on runner

Answer 39

sorry if i dont uderstand all this

Answer 40

Db on runner is the force of friction we calculated. The 176.6 N

Answer 41

Uh u mean runner on db?

Answer 42

Runner is applying 212 N, given in question.
Db is applying force of 176.6 N in opposite direction of force runner is applying.
Net force (the sum of these) is 35.6 N in the same direction of the force the Runner is applying.

Answer 43

At least that is how I would interpret the question/answer.

Answer 44

Ok im kinda confused, isnt the normal force times coefficient of friection=frcition force

Answer 45

because f(fric)=F(n)x(mew)

Answer 46

* mu

Answer 47

Yes, that is how we calculated the 176.6 N of force. That is the force of friction. It is opposite the applied force. So, the easiest way to visualize it is as the force the Db is applying on the runner (since that is in the opposite direction of the force the runner is applying to the Db).

Answer 48

When I described it as the force the Db was applying to the runner, that was just to give you a way to visualize that it was opposite the force the runner is applying.

Answer 49

yES I undertsand its opposite but ist not the applied force right?

Answer 50

No, the applied force is the one the runner is applying on the Db given in the question.

Answer 51

yes how do i find that?

Answer 52

is it equal to the force the db is applying on runner??

Answer 53

It is NOT equal to the force of friction. The applied force is given in the question. You don't need to do anything to find the applied force. The question gives it to you. It states "his applied force is 212 N"

Answer 54

nononono the applied force on runner is 212n, BUT what is applied force on Db from the runner. Do u undertsand what i mean?

Answer 55

This is all about wording. The applied force is 212 N. The 'force the runner is pushing the Db with' is (the question is poorly worded and this is just my opinion of what it really means) the net force. That is because it is actually the net force that is making the Db move (because the force of friction opposes the applied force )

Answer 56

So: Applied force = 212 N
Force of Friction was 176.6 N
Net force (or what I think the question means by 'the force the runner is pushing the Db with) is = 212 N - 176.6 N = 35.6N

Answer 57

Joe, the db is applying the force on runner, what im wondering is whats the runner applying. 212 is THE DBS force :)

Answer 58

Are you sure the Db is applying the 212 N force? I read it the other way around. I hate the wording of this. But, if it is the other way around....

Answer 59

yes

Answer 60

other way around

Answer 61

Db applying 212 N.
Normal force of Runner = 100 * 9.8 = 980 N
Force of friction = 0.2x 980 = 196 N

Answer 62

yes

Answer 63

and constant velocity

Answer 64

If the Db was the one applying 212 N, the force of friction (force runner applied to Db) would be 196 N.
The net force would be 212 - 196 = 16 N and the Db would be forcing the runner backwards

Answer 65

:o, so the applied force is the friction force also!

Answer 66

The applied force is OPPOSITE the force of friction in either case. They are not equal in either as I calculated above

Answer 67

but the applied force is 196n and friction is 196?

Answer 68

Where do you see 196 N for the applied force?

Answer 69

"the force of friction (force runner applied to Db) would be 196 N."

Answer 70

The question states: "his applied force is 212 N". Our only disagreement was on WHO was applying this force. We calculated it both ways if you scroll up you will see that.

Answer 71

.....

Answer 72

If the Db was the one applying 212 N, the force of friction (force runner applied to Db) would be 196 N.

Answer 73

Well even though this problem is worded weirdly. THANKS ALOT FOR YOUR HELP!

Answer 74

Really apprciate it