What is the coefficient of kinetic friction between the sled and snow?

[JonathanSnow]

[SOLVED] 2 more basic problems

3 dogs are pulling a sled with a combined weight of 145 kg the dogs are pulling
1) pulling at a force of 83N at 15.5(degrees) left of forward
2) pulling at a force of 75N at 9 (degrees) right of forward
3) pulling at a force of 77N at 12(degrees) right of forward

sled is moving a a constant velocity find the coefficient of kinetic friction between the sled and snow,
so the sled has a force of 1422.45N in the opposite direction the dogs are pulling, so then i have to find the vertical forces the dogs exert and add them together? so to find vertical i use f.sin(theta) then add the 3 together so i know.
1) Fy=22.18N
2) Fy=11.73N
3) Fy=16N

i don't even know if I am going about this the right way,

ill give problem 2 once i solve this

 

[JonathanSnow]

actually on thinking of it i was using Fg to find the Ff but now I am stuck as to how I am finding the Ff

 

[Mindscrape]

You are off to a good start, but be careful of which trig function you use. Now what does constant velocity imply in terms of acceleration and forces?

Ff = -µN

where N stand for the normal force.

 

[JonathanSnow]

constant velocity implies no acceleration which means net force is 0 but to find µ i need Ff do I not? did i do the forces of the dog right?

Fn would equal 1422.45 correct

 

[Mindscrape]

The forces from the dog are wrong because of a subtlety you missed, think about where the angles are defined from. Right, constant velocity implies no acceleration, so the forces from those dogs better cancel the force from friction.

 

[JonathanSnow]

so we can say Ff = whatever the F of the dogs combined is or the - of it, and i don't know what i missed on the forces is it supposed to be sin? or the angle part should i be starting my measurements at E ? anglewisethis gives me
1) force of 79.98
2) force of 74
3) force of 75

 

[Mindscrape]

Oh, well maybe you are right, but your numbers are wrong. Fdog*cosØ = Fforward

Yes, Ff = Fdogs because we know that Fdogs+(-Ff)=0.

 

[JonathanSnow]

so force dog cosØ is the same as the sin i mentioned when i edited my post when measuring off east? nice so FF=229 so i can go mu k equals 229/1422.45 which gives me .16 the book is saying .15x10(2) I am confused

 

[Mindscrape]

The dogs should have a total force of 80+74+75=150+80-1=229, might as well say 230 for niceness. You are right .16. What does your book say? 15? What??

 

[JonathanSnow]

yea the book says 15 but whatever if you agree with me that's all that matters.

 

[Mindscrape]

Well coefficients of friction don't exceed 1, so your book is screwy.

 

[JonathanSnow]

alright and the other problem i had was simple it is a pulley question where 2 blocks are connected block one is on the incline and weighs 47kg and block 2 is on the other end of the pulley and weighs 35kg, there's 4 parts but the only one i need help with is a will the blocks move if m2 is released, i know the answer is NO but how would you figure that out formula wise i should know this but worth asking

heres a diagram

red is Mass 1 Black is Mass2 and they move on a pulley, quite crappy 2 minute diagram

 

[JonathanSnow]

i know you have to see if fa is greater then Ff but i don't know what formulas to use..any help

 

[JonathanSnow]

anyone, id love to get this unit done with today, last question to do, still stumped and looked back through the textbook

 

[AstroRoyale]

Draw a free body diagram for block A. There is a force of m2*g pulling to the right, and to the left a force of m1*g*sin(25) from the weight of the block, plus whatever friction provides, ie mu*m1*g*cos(theta).

 

[JonathanSnow]

so i do all that and if they equal the same the system won't move correct? so i know Fn is 417.87 for block 1 and i know the right force is 343.35N so to find left components i can go Fleft + Ff so 194.85 +175.5 that equals 369 though which would mean it would move down the ramp

 

[AstroRoyale]

what is the coefficient of friction?

 

[JonathanSnow]

muk is 0.19 and mus is 0.42

 

[AstroRoyale]

for block 1:
m1*g*sin(theta)+mu*m1*g*cos(theta) - T = m1*a

For block 2:
T - m2*g = m2*a

eliminating T and solving for a
a=g*(m1*sin(theta) + mu*m1*cos(theta) - m2)/(m1 + m2)

The only way the system stays at rest is when
m1*sin(theta) + mu*m1*cos(theta) - m2 = 0

 

[JonathanSnow]

so my book is saying that it stays at rest? is it correct,

because that's not the answer i am getting

 

[JonathanSnow]

? no one i don't think it stays at rest, i think my books wrong can anyone comfirm this?

 

[JonathanSnow]

Could someone please confirm that there is actually movement if it let go it is not equaling 0. I'm doubting myself again

 

[JonathanSnow]

actually, it seems like the answer becomes 2.75kg which would mean in the direction of M1 which wouldn't be what we wanted so actually we could get it moving in the right direction by adding 2.8kg of weight to m2

does that make more sense then i can just figure out the acceleration of the combined weight of 37.8 which would mean the combined acceleration would be 1.14m/s(2) the book says its 5.7 but that's kind of unbelievable, anyone agree with 1.14?