Question

Duane has a mass of 73 kg and slides down an icy hill on a sled. The hill has a slope of 15° and there is no friction. The free-body diagram is shown below.

a. Write the expression for the net force in the x-direction.
b. Write the expression for the net force in the y-direction.

c. What is the acceleration in the y-direction?

d. What is the normal force acting on him?

Answer 1

Do you have the free body diagram?

Answer 2

first you have to resolve these forces into horiz/vertical components, keeping in mind the tilted axis

Answer 3

the component of gravity down the slide is wsin(15)

so overall we have:

net force in x-direction = Fg - Ff = wsin(15) - Ff
net force in the y-direction = Fn - Fg = Fn - wcos(15)

since a and b are just asking to set up the expression leave it like that

Answer 4

for c) set the y-direction expression equal to ma and solve for acceleration

d) set the x-direction expression equal to 0 since the box stays on the x-axis, then solve for normal force

Answer 5

BAck!

Answer 6

Okay a is Fg - Ff = wsin(15) - Ff

Answer 7

good

might wanna stick a ΣFy just to be claer

so ΣFy = Fg - Ff = wsin(15) - Ff

Answer 8

Fx sorry

Answer 9

other one would be ΣFy = Fn - Fg = Fn - wcos(15)

Answer 10

ΣFx = Fg - Ff = wsin(15) - Ff

Answer 11

for a right

Answer 12

ye

Answer 13

sorry just double checking

Answer 14

okay now for c

Answer 15

m*a right

Answer 16

f=m*a

Answer 17

yeah

I would actually recommend starting d) first since you need the normal force first

Answer 18

I wish we could edit our posts but you need to set the expression for y-direction force (not x) equal to 0 and solve for Fn

Answer 19

so Fn - wcos(15) = 0

where w is the weight (mg)
solve for normal force Fn

Answer 20

Fn - 715.4)cos(15) = 0

Answer 21

good, Fn = ?

Answer 22

691.02

Answer 23

good, so N

let's go back to a and b b/c I just re-read the problem and there's no friction so we need to get rid of the Ff term

Answer 24

net force in x-direction = Fg = wsin(15)
net force in the y-direction = same as before

Answer 25

a. Write the expression for the net force in the x-direction. (2 points)

Fg = wsin(15)
b. Write the expression for the net force in the y-direction. (2 points)
ΣFy = Fn - Fg = Fn - wcos(15)
c. What is the acceleration in the y-direction? (1 point)

d. What is the normal force acting on him? (2 points)
Fn - 715.4)cos(15) = 0
Fn=691.02N

Answer 26

good

to find the acceleration in the y-direction just set ΣFy = ma and solve for a

we determined w and Fn (normal force) earlier

Answer 27

Σ(691.02)y = (73)a

Answer 28

691.01 is the normal force not the net force

the net force in the y direction is Fn - wcos(15)

Fn - (mg)cos(15) = ma

Answer 29

691.01n - (715.4)cos(15)

Answer 30

hm. I'm getting weird numbers for this :S let me review my work

Answer 31

Sure, take your time

Answer 32

ugh of course, since the axis is tilted the box doesn't move up or down with respect to the tilted axis so acceleration is just 0 m/s/s

Answer 33

Got it so c is 0 m/s/s

Answer 34

yeah (tilted axes always mess me up since I typically use a horizontal/vertical axis)

Answer 35

OK thanks !!

Answer 36

Jake pulls a couch of mass 38 kg across his apartment. He pulls on the couch at an angle of 25° with a force of 300 N. The free-body diagram is shown below.


a. What is the x-component of his applied force?



b. What is the y-component of his applied force?


c. Assuming the couch does not leave the ground, what is the normal force acting on the couch?

Answer 37

x component = Fcos(theta)
y component = Fsin(theta)

(this is generally true for an object sitting on a flat surface, for tilted slopes we might have to think differently)

Answer 38

x component = 300cos(25)=271.9
y component = 300sin(25) =126.79

Answer 39

good, units are N as usual

for c) the net force in the y direction is 0 (box is not moving up or down)

so Fn + Fsin(theta) - mg = 0
solve for Fn

Answer 40

Fn + 300sin(25) - 38(9.8) = 0

Answer 41

good, solve for fn

Answer 42

245.61

Answer 43

N

Answer 44

yeah good

Answer 45

Mazie stands on her kitchen floor. The coefficient of kinetic friction between her socks and the floor is 0.35, and the coefficient of static friction is 0.42. She has a mass of 58 kg.

a. What is the maximum force of static friction between her socks and the floor?

b. If you push Mazie with a force of 200 N, will she start to slide? Explain your answer.

c. Mazie slides across the floor at a speed of 1.3 m/s. What is the force of kinetic friction acting on her?

Answer 46

58*0.42

Answer 47

mu = coefficient of friction (not mass * u)

static friction = normal force * mu

normal force = mg in this case

Answer 48

wait so for a i would multiply right

Answer 49

normal force = mg

static friction = μ*mg = ?

Answer 50

so mg=58*9.8

Answer 51

yes, keep going

Answer 52

578.2

Answer 53

good, keep going

Answer 54

static friction = μ*578.2 = ?

Answer 55

good, keep going

Answer 56

for u i would substitute one of the forces?

Answer 57

μ is the coefficient of static friction

Answer 58

0.42*578.2 = 242.844

Answer 59

good, 242.844 N

for b) since static friction = 242.844 is more than the applied force 200N she does ~not~ slide

Answer 60

for c) calculate μN except this time use the coefficient of kinetic friction for μ

Answer 61

0.35*242.844

Answer 62

good, then just evaluate that to get the ans

Answer 63

85

Answer 64

good, 85 N since this is a force

Answer 65

You slide a hockey puck across the ice in a hockey rink. The puck gradually slows down as it moves. How does Newton's first law explain the motion of the puck?

Answer 66

newton's first law - object doesn't change motion unless a force acts upon it - so there must be a force (friction) slowing it down

Answer 67

could you check my answer

Answer 68

In the case of a puck sliding across the ice: it slows down because of the friction force between ice and puck. This friction force is the external force spoken about in the 1st Law, above and is the reason the puck eventually stops moving.

Answer 69

awesome