# Friction problem with pulleys

1. Nov 15, 2005

The block with mass m2=0.50kg is found to have a speed of 0.30m/s after it has dropped 0.80m. how large a (kinetic) friction force retards the motion of the block with mass m1=2.0kg?

(side note from teacher)
To solve this problem, you can use a string constraint. the total length of the string (L) is constant. However, as block m1 moves to the right, the portion of the string that is horizontal shortens by delta y and this length is distributed over the portion of the string taht is vertical according to the relationship; |delta x|=1/2|delta y|. By taking a derivative with respect to time twice, we have a1=2a2.

i think i did this completely wrong

i drew a free body diagram for each,for 1st free body diagram (mass1) and found m1 had Fn of 19.6

on the 2nd free body diagram (mass 2), i had Fg-Ft = ma
then, 4.9 N - 2Ft =0 , Ft = 2.45, then i said the Ft in the first free body diagram was also equal to 2.45, and therefore the kinetic friction force equals 2.45

its probably wrong, since i didnt use any info my teacher gave in the side note

2. Nov 16, 2005

### Fermat

You put the accln of m2 as zero - it should be a2.

3. Nov 16, 2005

ohh.
okay, i did it like u said, but now im stuck

for the mass 1 fbd, i got
Ft - Ff = a1

for the mass 2 fbd, i got
Fg - 2Ft = ma2 (is it 2Ft???)
then isolate for a2
4.9 - 2Ft = 0.5a2
9.8 - 4Ft = a2

and the equaion from fbd #1:
Ft - Ff = ma1
since a1 = 2a2
Ft - Ff = 4a2 (1)
then i substituted a2 into equation (1)
Ft - Ff = 4(9.8 - 4Ft)
Ft - Ff = 39.2 - 16Ft
17Ft = 39.2 + Ff

okay im stuck, how do i find Ft???

4. Nov 16, 2005

### Fermat

You have,

9.8 - 4Ft = a2 (from fbd #2)
Ft - Ff = 4a2 (from fbd #1)

eliminate Ft from these two eqns, giving you,

9.8 - 4Ff = 17a2

Now use the info about the movement of m2 given in the question to work out a2.
Substitute for a2 and solve for Ff.

5. Nov 16, 2005

so, a2 = 0.30m/s / 0.80m = 0.375 m/s^2
and then sub it in to get Ff = 0.856 N
oops, thats not the acceleraton
hmm, u need kinematics??
is the veloctiy given v1 or v2
a = 9.8 m/s^2
d= 0.80
if velocity given is v2, so v1 = 0?

Last edited: Nov 16, 2005
6. Nov 16, 2005

actually i have a question, i used 2Ft in one of the equations, and im wondering if this is right.

for this step i used 2Ft :
for the mass 2 fbd, i got
Fg - 2Ft = ma2 (is it 2Ft???)

7. Nov 16, 2005

### Fermat

Yes, that's right.

Nope, that's wrong
use the kinematic eqn,
v² = u² + 2as