Quick look over friction pulley system

[ngorecki]

Homework Statement

Objects with masses m1 = 12.0 kg and m2 = 4.00 kg are connected by a light string that passes over a frictionless pulley. If, when the system starts from rest, m2 falls 1.00 m in 1.70 s, determine the coefficient of kinetic friction between m1 and the table.

Homework Equations

Fg = Fn = mg
F = ma
μ = Ff/Fn

The Attempt at a Solution

Fg = mg
Fg = 16*9.8
Fg = 156.8 = Fn

F = ma [m2]
= 4*9.8
= 39.2

ƩFx = m2 - Ff (sum of all forces in the x direction is = m2 - Frictional force)
= 39.2 - Ff
Ff = 39.2

μ = Ff/Fn
μ = 39.2/156.8
μ = .25

I tried entering the coefficient of kinetic friction into the "submit box" but the answer is incorrect.
could somebody please point out my mistake?

 

[ehild]

Is there a table in the problem? A drawing would be nice, with free body diagram for both masses.

ehild

 

[ngorecki]

this is the picture i have

 

[ehild]

Nice picture.
Both blocks accelerate with the same amount as they are connected with a spring.
m2 falls 1 m in 1.7 s. What is its acceleration?

ehild

 

[ngorecki]

thanks :)
and acceleration is:
Δx = .5 a t^2
1 = .5(a)1.7^2
1 = 1.445a
.69 = a

 

[fLambda]

Your mistake is in your calculation of normal force. Only m1 should factor into normal force. M2 experiences only gravity and tension :)

 

[ngorecki]

i multiply m1 by .69?

 

[ehild]

It is OK,
but do not round off during the calculations, keep one more digit as the number of the significant digits in the data. And show the unit. So a=0.6920 m/s²

The acceleration of m₂ is know, so the net force on m₂ is F=m₂a. What forces act on m₂?

ehild

 

[ngorecki]

gravity and tension?

 

[ehild]

i multiply m1 by .69?

If you multiply m₁ with the common acceleration, you get the net horizontal force on m₁. That force is the resultant of what forces?

Think of the string. Does it exert force?

ehild

 

[ngorecki]

the string which is connected to m2 pulls m1. and then there is friction on m1

 

[ehild]

gravity and tension?

Yes. Gravity and tension of the rope result in F=m₂a. Can you find the magnitude of the tension?

ehild

 

[ehild]

the string which is connected to m2 pulls m1. and then there is friction on m1

Yes, what are the direction of these forces?

ehild

 

[ngorecki]

gravity on m2 = 39.2 N. the tension = 39.2 also?
and friction is in the "negative" direction
force from m2 is in the "positive" direction

 

[ehild]

gravity on m2 = 39.2 N. the tension = 39.2 also?
and friction is in the "negative" direction
force from m2 is in the "positive" direction

The tension in the rope acts upward on m₂. See the picture, I added the forces. So the net force on m₂ is m₂g-T=m₂a. What is T then?

ehild

 

[ngorecki]

36.432 = t

 

[ehild]

36.432 = t

Yes, the tension is T=36.432 N.
Now what forces act on m₁?

ehild

 

[ngorecki]

Tension and friction?

 

[ehild]

Yes. The tension is the same at both ends of the string. You know acceleration, mass and one force, tension. How much is the other force, the force of friction?

ehild

 

[ngorecki]

ƩFx = T - Ff
ƩFx = 36.432 - Ff
Ff = 36.432 N

 

[ehild]

Is ƩFx =0? m1 is accelerating!ehild

 

[ngorecki]

M1 = ma
M1 = 12 * .692
M1 = 8.304

M1 = T- Ff
8.304 = 36.432 - Ff
Ff + 8.034 = 36.432
Ff = 28.398

Is this accurate?

 

[ehild]

Why do you denote force by M?

So the net force on m₁ is m₁a=8.304 N, and it is equal to T-Ff. But what is 8.034?

ehild

 

[ngorecki]

the mass of m1 * acceleration = 8.304
then i take the mass of m1 * gravity = 117.6
u = Ff/Fn = 28.128/117.6 = .239

the answer is.239

 

[ehild]

Correct !

ehild

 

[ngorecki]

Thank you very much!