# Homework Help: 3 Pulleys System, find the acceleration of both mass.

1. Jul 1, 2012

### Bonten

1. The problem statement, all variables and given/known data
Here is a picture that I putted together(see attached picture).
m1= 0.235 kg
N = m1g = 2.303 N
m2= 0.350 kg
m2g = 3.430 N
fk= 0.590 N

By measurement, I confirmed (d1)=4(d2).
Therefore, (v1)=4(v2), a1= 4(a2).

2. Relevant equations
ƩFx=(T1)-fk=(m1)(a1)
ƩFy=(m2)g-(T2)/2=(m2)(a2)
(T2)-4(T1)=0

3. The attempt at a solution
Try to find a1 and a2.
[(m2)g-(m2)a2)]/2-fk=(m1)4(a2)
1.715-0.175(a2)-0.590=(.235)(4)(a2)
1.125=1.115(a2)
(a2)=1.01 m/s^2
(a1)=4.036m/s^2

please correct me if I am wrong. Thank you.

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2. Jul 1, 2012

### ehild

Hi Bonten!

The expressions in red are not right. Think them over.

ehild

3. Jul 1, 2012

### Bonten

I got hang up on this all weekend.
1)would it be ƩFy=(m2)g-(T2)=(m2)(a2) ?
2)would (T2)-2(T1)=0 ? since T2 exerted the force from (m2)(g2) to the pulley (P1) which divided the tension by 2.
3)but the T2 got divided by 2 by the pulley P2 which directly attached with m2. here is the diagram, it applies with p1 pulley as well.

4. Jul 1, 2012

5. Jul 1, 2012

### ehild

No. Two pieces of string act upward, each with tension T2. m2g-2T2=m2a.

Yes, the pulley is massless so the sum of forces acting to it must be zero. 2T1 acts to the left and T2 acts to the right: T2-2T1=0.

Your diagram applies for equilibrium, when W-2T=0, so T=W/2.

ehild

6. Jul 1, 2012

### azizlwl

It can be seen clearly in your diagram, 2×T2 is opposing the motion of M2.

7. Jul 1, 2012

### Bonten

this is system not quite in equilibrium, since the downward motion of m2g will create an acceleration for the whole system, but a2 will be 4 times smaller than a1.