I think my professor just gave me some bullshit that doesnt make

the friction coefficient would have to be 6 for this shit to have a constant velocity

help please, my math anus is bleeding

This is a slow board you idiot, stop bumping.

>professor

That's a high school level problem. Kys brainlet.

problem is set in australia.

>>8721640
i never said i couldnt solve it, i said that the numbers dont work and this is a problem is made to fail in the firstplace.

>>872164this says the its moving down the incline but the hanging mass on the pully is 54 kg, while the object is 10, do you realize how retarded that statement is. there is no way this could slide downwards at all

>>8721647
I think he got the numbers for M and m mixed up.

>>8721665
thats what i was thinking till you realize that that these types of problems dont even get a constant velocity.

>>8721666
Yes they do.

>>8721612
>>8721675

>>8721689
im not asking for the answer, i know how to solve these problems. i just dont like getting problems from professors that are obviously flawed

File: d4b5997a88dcf05474e8276de4f5a284.png (78KB, 556x562px) Image search: [Google]

78KB, 556x562px

>>8721689
>>8721640
soo tell me how this isnt borked

>>8721706
stop being autistic and say to the guy "i think you got the numbers the wrong way round" jesus christ, you'd think this was his magnum opus the amount you are complaining.

>>8721706
I'm to lazy to check the baby tier math, but that online calculator probably uses static friction (gives you the initial acceleration when released from rest).

In your profs question, it's kinetic friction, so both masses are already moving at constant speed...

>>8721714
he sent this out to everyone as a study guide. i cant just walk up to him and say. yea you just sent over 300 students a broken problem that you told them to study for mondays exam

>>8721717
those weights just don't make any sense.

>>8721719
yes you can. it's called an e-mail. "oh hi professor, I think you may have mixed up the values" usually accompanied by "no, it's correct you brainlet" or "oh my bad fixed version up." that he'll email to everyone

>>8721728
I'm not OP, but even when you mix around the values it comes out with the wrong answer. The general formula used is u=(Msin(theta)-m)/(Mcos(theta)), and nothing you plug in comes even close to the values given.

>>8721728
in a class where the average dictates the curve im not going to be helping everyone else

>>8721612
Don't complain, see it as a challenge. Since M is supposed to move down the incline, it cannot be smaller than m. Assume M=54kg and m=10kg.

m=Msin(30)-µMcos(30)
m/M=sin(30)-µcos(30)
µ=(m/M-sin(30))/-cos(30)
µ=0.36351683615643104

So C) 0.4 is the answer.

>>8721612
Well, assume a new "dark" force.

>>8721612
Why would you make it an ABC problem?
Besides no rational answer can be correct since the problem involves an irrational number(cos30)

>>8722503
>what is rounding

>>8722015
but its not, the answer is clearly .1 as you see the one that is bold

THIS BOARD IS 18+ ONLY YOU LITTLE MOTHER FUCKER.

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>>8722587
Have you no mind of your own? It's the reasoning that counts, not the possibly forged image. Don't play dumb. Write a short assessment and send it to your instructor.

I am a math major, but can still answer this. Since velocity is constant, there is no acceleration acting on the system. So Mgsin(theta)=(Coefficient of kinetic friction)*mgcos(theta). The Kinetic friction is equal to (M/m)*tan(theta). Which gives .1, the answer A.

>>8721728
You are acting like this is some highly established scientific comity when its just a fucking board on 4chan

>>8721706
>>8721612
The smallest coefficient of kinetic friction that allows this scenario to work is around 5.658031

>>8723691
Because the system isn't accelerating, we can assume the tension in the rope cancels out with the force of gravity acting on the 54kg mass, so the tension should be 54*g. The incline is 30 degrees from the horizon, so gravity is pulling on the 10kg mass with a force of 10*g*cos(60) = 5g. The sum of the force vectors on the 10kg mass must be 0, as is it is not accelerating; take x to be the smallest possible coefficient of kinetic friction satisfying the problem, then we know that 54*g - 5*g - x*(10*g*cos(30)) = 0. So 49*g = (10*g*cos(30))*x, x = 49/(10*cos(30)) = 49/(10*sqrt(3)/2) = 49/(5*sqrt(3)).

>>8723656
Now we know that the author of OP's problem is a math major who has lost contact to reality.

File: yup.jpg (3MB, 2432x4320px) Image search: [Google]

3MB, 2432x4320px

>>8723691
>>8721612
Agreed, this question's fubar. 0.1 shouldn't be marked as the correct answer.

>>8721750
>needing to use tactics like this to protect your grade
brainlet detected

>>8723656
so essentially, professor forgot that acceleration due to gravity is lower for the block on the incline

>>8724802
I did fast calcs to corroborate you:

+x:
529.2 N

-y:
84.87 N

+y:
84.87 N

-x:
49 N

-----------------
+x > -x
it has to slide up

coefficient = (529.2-49)/84.87 = 5.6580652763

>>8721666
The frictional force is equal to the component of the force of gravity on M in the plane of the slope, plus the force of gravity acting on m. Forces cancel out, so there's no acceleration, but you could still push one and start motion.

>>8725169
Changing mases (M for m):

+x:
98N

-y:
458.30 N

+y:
458.30 N

-x:
264.6 N

-----------------
+x < -x
it has to slide down

coefficient = (264.6-98)/458.30 = 0.36351734671

Kill your sensei desu.

>>8725190
You are the only one being pushed off here.

>>8725195
What?

>>8725190
That's only a tactic to set the direction of frictional force vector.

>>8725200
It still explains why there would be a lack of acceleration. I wasn't trying to solve the problem, I was only trying to explain that a block in a system like that could be in motion without acceleration.

>>8725191
see >>8722015