Question

A sled of mass 50 kg is pulled along snow-covered, flat ground. The static friction coefficient is 0.30, and the sliding friction coefficient is 0.10.
---
a) What does the sled weigh?
b) What force will be needed to start the sled moving?
c) What force is needed to keep the sled moving at a constant velocity?
d) Once moving, what total force must be applied to the sled to accelerate it 3.0 m/s/s?
---
Answers:
a. 490 N
b. 150 N
c. 49 N
d. 200 N

Answer 1

Firstly, I need to make sure the coefficient variables are right.
Static friction:\[\mu _{s}\]
Sliding friction (I take this means kinematic?):\[\mu _{k}\]or\[f\]

Answer 2

phew. okay, sleds.

and yeah, sliding friction would be kinetic friction.

Answer 3

coefficient of kinetic friction is
\[\mu_k\]

Answer 4

Okay, gotcha.
Is this right?\[\mu _{k} < \mu _{s}\]

Answer 5

yah

Answer 6

... Now I'm lost...

Answer 7

^_^ So first, what does the sled weigh, and what is the summation of forces for the first part (static friction)?

Answer 8

Part A\[w = mg = 50 \times 9.8 = 490 N\]
Part B\[F = \mu N\]... I don't know what to do now..

Answer 9

That second part is the summation of the forces in the x-direction; and this is for part B, so make sure to label that mu as mu s.

For an object on a horizontal surface, what is its normal force?

Answer 10

Part B\[F = \mu _{s}N = 0.30N\]

Answer 11

I don't know either F or N..? o.o

Answer 12

for an object on a horizontal sutface, the Normal force is equal to its weight; you know that the force it takes to start the object moving is when

\[F≥\mu_sN\]

So yeah, just like you wrote
\[ F≥(.3)N\]

Answer 13

... GRAH. We did NOT learn this like that = n =;;

Answer 14

just use equals if it makes more sense ^_^

Answer 15

you're looking for the minimum force it would take, so it's an "equals' anyway

Answer 16

Well it's in the notes ... but used this for in-class problems:\[F _{A} = f = \mu N\]

Answer 17

We also used the other one with an = sign.

Answer 18

whatever makes the most sense ^^

Answer 19

But when I learned it in the NOTES:\[-F _{f} \le \mu _{s}N\]

Answer 20

Or Fn as they say, but same thing right?

Answer 21

\[F_n=N\]
yah

And yeah, the force of static friction will be less than or equal to mu s N

Answer 22

What about this?

Answer 23

you mean why did I write greater than?

Answer 24

\[\Sigma F_{y} = N - W = ma\]ma cancels out to 0\[\Sigma F _{y} = N - W = 0\] W = 490 (leaving N out because there is a variable N)\[N = W = -490\]

Answer 25

Sorry I should have typed that all together.

Answer 26

N=+490 N

Answer 27

Yeah sorry I messed that up lolol

Answer 28

then you is correct in what you wrote ^^

Answer 29

WOOT!
But how do I find Fa?

Answer 30

OWAIT!\[\mu _{k} = \frac{ F _{A} }{ N }\]\[0.10 = \frac{ F _{A} }{ 490 }\]\[F _{A} = 49\]
I feel stupid now = n =;;

Answer 31

but you figured it out - you should feel smart! ^_^

If you set that up with a force summation you get the same thing. When the sled is moving, it's moving at a constant velocity, so a=0 as well

\[ \sum F_x = F_A - F_{kinetic friction} = 0 \ \ \longrightarrow \ \ F_A = \mu_kN\]

Answer 32

Got the last bit figured out?

Answer 33

HALLO?? @kittiwitti1

Answer 34

Oh um ehh I was afk? xD

Answer 35

figured you had fallen asleep at the keyboard ^_^

Answer 36

not that =_="
I actually took a nap beforehand

Answer 37

Wait - why do you even need the first part of the equation?

Answer 38

you used the definition of the coefficient of kinetic friction - the way I did it derived it from newton's second law. They're equivalent ^_^ Is that what you meant?

Answer 39

No, I meant why do you even have to type it out...

Answer 40

**Write it out

Answer 41

habit? I don't get what you mean...

Answer 42

oh! You were using FA to be the actual force of kinetic friction! I get it. I thought FA was the force being applied to the sled.

\[F_{push} - F_{k \ friction} = 0 \longrightarrow\quad F_{push} = F_{k \ friction} \longrightarrow \quad F_{push} =\mu_kN\]

Answer 43

Wait WHAT?! NO!! o_o
I'm using Fa for applied force!!

Answer 44

Then I really don't understand your question... :/

Answer 45

@ _ @ I don't get it anymore either

Answer 46

you have the right answer...

Answer 47

Is the front part of the equation necessary to write out*

Answer 48

If you say it's newton's second law yes; if you say it's the third, no, since there are only two forces in this system

Answer 49

We learned to use this formula:\[F = \mu _{k}N\]

Answer 50

That's the definition for the force of kinetic friction, yah, and setting it equal to "force applied" works in this instance because it's the only other force in the problem

Answer 51

We didn't learn the first part...

Answer 52

I mean, F kin. friction? What is that o_o

Answer 53

Everything you do until forever in mechanics always relates back to Newton's Second law. In this problem, the force opposing the motion of the sled is friction, so you can always set up a force diagram and sum the forces like what I typed out up there.

Answer 54

Is it okay to skip it?

Answer 55

WAIT do you mean Fa - f = ma?

Answer 56

yesyesyesyesyes

Answer 57

OHHHH lols wow I'm so stupid xD

Answer 58

\[ \sum F_x = F_{applied} - F_{friction} = \cancel{ma}^0\]

Answer 59

How did you cross that out o.o

Answer 60

\cancel{ma}^0

Answer 61

no on the console what's the button for it

Answer 62

there isn't one - I don't use the console

Answer 63

WOW. tech genius lol

Answer 64

\sum F_x = F_{applied}-F_{friction} = \cancel{ma}^0

waaaay faster

Answer 65

oh, I meant I don't use the buttons

Answer 66

Exactly, so you figured that out on your own! :p
\[Sum F_x = F_a - f = \cancel{ma}^0\]
I did "[Sum F_x = F_a - f = \cancel{ma}^0"

Answer 67

Thanks for the tip~

Answer 68

welcome -
\[ \LaTeX\] is wonderful!! you should learn it

Answer 69

every operation has to have \ before it, like \sum

\[\sum\]

then

\[

at the beginning of the phrase and

Answer 70

\]

at the end

Answer 71

\[ \text{ \[ \sum F = ...... = \cancel{ma}^0\] }\]

Answer 72

needs \

Answer 73

the symbol right below the delete key

Answer 74

three right of "P"

Answer 75

in any case, the last part isn't too bad

\[ \sum F_x = F_{applied} - F_{k \ friction} = ma \quad ; \quad a=3m/s^2
\\ \
\\ \
\\ \
\qquad F_{applied} = ma + F_{k \ friction} = ma + \mu_k N\]

Answer 76

O_O wat

Answer 77

another Newton's second law

Answer 78

The net force on the sled is
\[ma\]
, and the two forces that are acting on it are
\[F_{applied}\]
and
\[F_{kinetic friction}\]

Answer 79

ok I'm really confused now, are we doing c) or d)? = n =;;

Answer 80

So to have the sled accelerate at 3m/s^2, the force applied to the sled has to overcome the force of kinetic friction, and also accelerate the sled at 3 m/s^2 - therefor, the force applied must equal the force of kinetic friction plus the mass of the sled times its acceleration (3m/s^2)

Answer 81

Oh, this is for the last part. I'm sorry. I thought you had already done C...

Answer 82

I'm not done with c)

Answer 83

for the part a of the question you were right
for the second part, you should just find the normal force between the block and the ground and then multiply it with the coefficient of static froction to get the limiting condition this means that if you apply force of magnitude greater than this static friction, the sledge will start moving
next for part c
as you have dine it before, that total forces on the sledge should be zero, so it will move with zero acceleration so, it will have constant velocity.
but this time, the block is moving so, you should multiply the coeff. of kinetic friction(specified as sliding friction) with the normal force to get the frictional force
next for the part d, f=ma
f=50x3=150 and a force of 49 n is already acting on the block to make it move with a constant velocity.
thus by adding both the forces i.e. fand 49n, we will get the total force=199n nearly equal to 200n

Answer 84

You wrote the answer down up there

\[F_A = \mu_kN = (.10)(490N) \\ \\ \
F_A = 49N\]

Answer 85

A sled of mass 50 kg is pulled along snow-covered, flat ground. The static friction coefficient is 0.30, and the sliding friction coefficient is 0.10. --- a) What does the sled weigh? b) What force will be needed to start the sled moving? c) What force is needed to keep the sled moving at a constant velocity? d) Once moving, what total force must be applied to the sled to accelerate it 3.0 m/s/s? ---

Answers:
a. 490 N
b. 150 N
c. 49 N
d. 200 N

Answer 86

.... nvm

Answer 87

I misread...

Answer 88

Wait, then how to get b)?

Answer 89

\[F_A=\mu_sN\]
\[F_A=(.3)(490N)\]

Answer 90

NEVERMIND! I got it.

Answer 91

^_^

Answer 92

ok what's d)?

Answer 93

from @rajat97
next for the part d, f=ma
f=50x3=150 and a force of 49 n is already acting on the block to make it move with a constant velocity.
thus by adding both the forces i.e. fand 49n, we will get the total force=199n nearly equal to 200n

Answer 94

idk, it's confusing

Answer 95

Write out the forces acting on the sled

Answer 96

idk
my homework is due in less than an hour and I have 5 problems left -_-;;

Answer 97

wait Fa(k) + Fa(s)?

Answer 98

I don't know what that means