Question

Blaine steps onto a ski slope with an angle of 25°. There is a coefficient of kinetic friction of 0.15 between him and the ground. He has a mass of 65 kg.

a. Draw the free-body diagram showing the forces acting on Blaine.

b. Write the expressions for net force in the x- and y-directions. Be sure to tilt your axis along the incline.

c. Calculate the net force and acceleration in the x- and y- directions.

Answer 1

@Vocaloid

Answer 2

I"m gonna assume he's going down the slope

Answer 3

by "tilt the axis along the incline" we are going to define our coordinate system that puts Fn along the y-axis and mg at an angle

Answer 4

okay so is this the complete drawing for A

Answer 5

yes

Answer 6

i would assume it is going down as well

Answer 7

now is the difficult part, re-solving mg into it's horizontal/vertical components

Answer 8

so just looking at the gravity in the y direction gives us

Answer 9

making the horizontal component sin(35)mg

Answer 10

okay so from what you describes it all boils down to solving for the horizontal component of sin(35)mg

Answer 11

yeah, the tricky part is getting that 35 by using the properties of a line (all angles add up to 180 for a straight line, right angles are 90, etc)

Answer 12

using the angle that is given of course

Answer 13

right, so i would solve horizontal component sin(35)(65)(9.8)

Answer 14

okay so for b its just

Fy = Fn - mgcos(35)
Fx = Fn - mgsin(35)

Answer 15

no answers just expressions?

Answer 16

oh right, since she's sitting still the net force in both directions is 0

so Fy and Fx are 0

but yeah they just want expressions not any values

Answer 17

and of course I forget the friction force ugh

Answer 18

0 = Fn - mgcos(35)
0 = Fn - mgsin(35)

Answer 19

sorry the y-direction is right but the x direction needs to be

0 = mgsin(35) - Ff since there's no normal force in the x direction

Answer 20

actually they're probably not equal to 0 ugh just leave it as Fx and Fy

Answer 21

Fy = Fn - mgcos(35)
Fx = mgsin(35) - Ff

that should do it

Answer 22

okay so for c

Answer 23

i'm confusing myself give me a minute

Answer 24

Sure take your time

Answer 25

ok, since the net force in the y-direction is 0 (the tilted axis was confusing me but he doesn't move up or down the y-axis)

so normal force = mgcos(35) = ?

Answer 26

(65)(9.8)cos(35)

Answer 27

good, then multiply that out and stick a N unit on it

Answer 28

521.8

Answer 29

N

Answer 30

good, that's the normal force which will be used in both the Fx and Fy calculations

Fy = Fn - mgcos(35) = ?

(plug in normal force for Fn)

Answer 31

Fy = 521.8 - (65)(9.8)cos(35) = ?

Answer 32

0.00015

Answer 33

ugh I"m an idiot we already said that Fy is equal to 0

so yeah Fy = 521.8 - (65)(9.8)cos(35) = 0

(the extra 0.0015 is just because of rounding, in theory it should be 0)

Answer 34

anyway, Fx = mgsin(35) - Ff

where Ff = friction force = coefficient of friction * normal force

Answer 35

21.8 - (65)(9.8)cos(35) = 0

Answer 36

so am i solving for anything here

Answer 37

we already determined it was equal to 0, so we just state that net force in the y direction is equal to 0 N

Answer 38

next step is to calculate Fx which is just Fx = mgsin(35) - Ff

Answer 39

Fx = mgsin(35) - Ff

Answer 40

where Ff is the friction force which is coefficient of fric * normal force

Answer 41

521.8*0.15=78.27N

Answer 42

goood, so mgsin(35) - Ff = ?

Answer 43

65(9.8)sin(35)-78.27=287.1N

Answer 44

good so far we have

Fy = 0
Fx = 287.1N down the slope (defining down the slope as the + x direction)

it also asks for acceleration in the y and x directions, and according to newton's 2nd, F = ma

so acceleration is just F/m, just divide both forces by m

Answer 45

521.8/65

Answer 46

8.03N

Answer 47

good,

however we are solving for acceleration not force so m/s/s are units

Answer 48

acceleration in the y direction is just 0 m/s/s since 0/65 = 0

Answer 49

is it okay if i just show you what i wrote for C

Answer 50

sure

Answer 51

Calculate the net force and acceleration in the x- and y- directions.

Normal force = mgcos(35)
(65)(9.8)cos(35)=521.8N
Net force in the y direction is equal to 0 N

521.8*0.15=78.27N
65(9.8)sin(35)-78.27=287.1N

Acceleration is just F/m
521.8/65=8.03m/s/s

Answer 52

good just let me be a bit more specific w/ certain steps\(\color{#0cbb34}{\text{Originally Posted by}}\) @zarkam21
Calculate the net force and acceleration in the x- and y- directions.

Normal force = mgcos(35)
(65)(9.8)cos(35)=521.8N
Net force in the y direction is equal to 0 N

521.8*0.15=78.27N ---> specify that you are solving for friction here
65(9.8)sin(35)-78.27=287.1N ---> specify that this is the net force in the x direction

Acceleration is just F/m
521.8/65=8.03m/s/s ---> specify that this is the x direction only
also include the fact that acceleration in y direction is 0
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 53

GO it !! Thank you so much !

Answer 54

oml i'm actually going to punch a wall

the angle should be 25 not 35 so just repeat the calculations with 25 ;;

Answer 55

Normal force = mgcos(25)
(65)(9.8)cos(25)=577.32 N
Net force in the y direction is equal to 0 N

521.8*0.15=86.598 ---> specify that you are solving for friction here
65(9.8)sin(25)-86.598=182.61N ---> specify that this is the net force in the x direction

Acceleration is just F/m
182.61N/65=2.809 m/s/s ---> specify that this is the x direction only
also include the fact that acceleration in y direction is 0


^ there that should be it

Answer 56

for b) just replace the angles

Fy = Fn - mgcos(25)
Fx = mgsin(25) - Ff

everything else is the same

Answer 57

and for a) it's just like this

Answer 58

oh you beat me to it! I was just doing it as well