Force and Kinetic friction problem

[bmandrade]

A sled weighing 100 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is .275. A penguin weighing 50 N rides on the sled

to start with the penguin digs in his claws so he is firmly attached to the sled. What value of F do you need for the sled and penguin to move at constant speed??

after a while, our penguin gets tires od holding on with his claws. Now the coeffient of static friction between penguin and sled is .750 find the maximun horizontal force F that can be exerted on the sled before the penguin bgins to slide off?

 

[tiny-tim]

Hi bmandrade!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[bmandrade]

Ok well I've been doing some diagrams
In my diagrams for 1st part I show that there is a force pulling the sled and opposite to that there is kinetic friction. Force due to gravity is pointing down and a normal pointing up.
for the penguin I show a force which is opposite to the one that pulls the sled. Fg downwards and a normal upward.

For sled
I found the normal to be 100 N since its the same as Fg which 100 N. I multiplied this number times the Kinetic friction to find the force due to kinetic friction (100 N * .275 = 27.5) I knwo that the sum of forces on the horizontal direction are going to equal ma
and mass is 100N/g = 10 Kg (g=10m/s^2)
so
sum of all forces in x = F - 27.5 = 10kg (a)

now I am stuck

 

[tiny-tim]

What value of F do you need for the sled and penguin to move at constant speed??
…
For sled
I found the normal to be 100 N since its the same as Fg which 100 N. I multiplied this number times the Kinetic friction to find the force due to kinetic friction (100 N * .275 = 27.5) I knwo that the sum of forces on the horizontal direction are going to equal ma
and mass is 100N/g = 10 Kg (g=10m/s^2)
so
sum of all forces in x = F - 27.5 = 10kg (a)

Excellent …

and constant speed means the acceleration, a, is … ?

 

[bmandrade]

acceleration = 0

 

[bmandrade]

does that mean that the minimum value for the sled to move will be 27.5 N?

 

[tiny-tim]

… poor little penguin!

does that mean that the minimum value for the sled to move will be 27.5 N?

Yes … except you forgot the penguin!

 

[bmandrade]

ok... I don't know how to deal with the penguin

 

[tiny-tim]

ok... I don't know how to deal with the penguin

Feed it fish, of course! ​

I meant, you forgot its weight!

 

[bmandrade]

lol


mmmm i not sure but do I add penguins weight to the force?

 

[tiny-tim]

You add the penguin's weight to the sled's weight to calculate the normal force, to calculate the friction force.

 

[bmandrade]

oh so its the weight of the whole system
in that case the Force = 41.25 N
becasue Force due to friction is 150(.275) = 41.45 N


so then I use this force to solve part 2 right?

if I am correct then the Static friciton between penguin and sled is
(.750*50)= 37.5 N

so the sum of forces in x direction = force of kinetic friction - force= 0?

 

[tiny-tim]

oh so its the weight of the whole system
in that case the Force = 41.25 N
becasue Force due to friction is 150(.275) = 41.45 N

Yes.

so then I use this force to solve part 2 right?

No … it's a different F … you start again.

if I am correct then the Static friciton between penguin and sled is
(.750*50)= 37.5 N

Yes.

so the sum of forces in x direction = force of kinetic friction - force= 0?

Nooo … for the penguin, the sum of forces in x direction = force of kinetic friction = mass x acceleration.

 

[bmandrade]

so for part 2 the force will be different is that force going to be 27.5 or no??

and for the last part where you said no i don't know how to do that because there is no acceleration

 

[tiny-tim]

so for part 2 the force will be different is that force going to be 27.5 or no??

No … the force is unknown … it's the (different) F which you have to calculate.

and for the last part where you said no i don't know how to do that because there is no acceleration

ah … yes there is … the penguin won't come off without acceleration, will it?

 

[bmandrade]

ok so can you help me figure this out because i have no clue how to do it

 

[tiny-tim]

ok so can you help me figure this out because i have no clue how to do it

The penguin will start to slide when the acceleration is enough to balance the maximum possible friction force …

so that gives you the acceleration of the penguin …

the acceleration of the sled-plus-penguin is the same (just before it starts to slide) …

so that gives you the force F on the sled-plus-penguin.

Have a go!

 

[bmandrade]

sorry but i still don't get it

this is what i have
the Fs= 37.5 N
the sum of F is x = Fs - F = ma

but i don't know what F is so i can't solve for a

 

[tiny-tim]

the sum of F is x = Fs - F = ma

No (and what's x? ) … it's only Fs = ma … the friction is an internal force, between the penguin and the sled …

so the friction will only show up in a Newton's second law equation for either the penguin on its own, or the sled on its own …

in this case, do the penguin on its own, to find a, then do the penguin-plus-sled, to find F.

 

[bmandrade]

ok so

the penguins acceleration will be .75 m/s^2
i used fs=ma so 37.5 = 50a -------> a = .75 m/s^2


and i use this value for F=ma

which will be F = (.75m/s^2) * 150N = 112.5?

 

[tiny-tim]

the penguins acceleration will be .75 m/s^2
i used fs=ma so 37.5 = 50a -------> a = .75 m/s^2

No … you've used m twice (they'll cancel), and you haven't used g.

 

[bmandrade]

ok sorry but i just don't get it maybe if you explain it with numbers will be easier to understand

 

[tiny-tim]

ok sorry but i just don't get it maybe if you explain it with numbers will be easier to understand

normal force = 50

mass = 50/g

Fs = 50µ = 50a/g.

 

[bmandrade]

oh ok
so this gives a to give 7.35 m/s^2

so i used this number for f=ma

which is f= 150(7.35)/g = 112.5

 

[tiny-tim]

oh ok
so this gives a to give 7.35 m/s^2

so i used this number for f=ma

which is f= 150(7.35)/g = 112.5

Yes!

 

[bmandrade]

ugh Thank you so much it was a long and confusing problem.