# System of Masses and Pulley, Problem

## Homework Statement

Calculate T1, T2 and T3. JPeg image of problem is attached.

## Homework Equations

Sum of all forces = Mass of system x Acceleration of System

## The Attempt at a Solution

I think I have T1 and T3 figured out but I am having problems with T2.
if the (E = sum of)
EF = ma
T2 + Ff = ma (Ff = 0 b/c u = 0)
however 0.5 kg is hanging so Fg is acting on T2
So i treated Fg of 0.5 kg mass like I would friction. The thing that is troubling me is that I thought you could only add up forces in the same plane. i.e. x-plane or y-plane.

a = 0.61 m/s^2 to the left, from previous work

How do I relate Fg from the 0.5 mass with T2 (going left?)

My first attempt:

T2:
EF = ma = T2 +Fg
(4.5 kg)(0.61m/s^2) = T2 + (0.5kg)(-9.8m/s^2)
T2 = 7.6 N

I had two issue. B/C everything is accelerating to the left, I thought a should equal -0.61m/s^2 but then the T2 value seems too small.

The second issue relates to Fg in the y-plane and T2 in the x-plane. Can you sum these forces up?

Should t1 > t2 > t3??? or not?

Thanks

#### Attachments

• System of Masses Problem.JPG
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Last edited:

ehild
Homework Helper
Gravity acts on every masses, but the gravitational force of one mass does not act on an other mass. Consider both blocks on the table separately. What forces act on the 2.5 kg block? Gravity of magnitude 2.5g acts downward, but it is cancelled by the normal force from the table. Two pieces of string are connected to it. One piece pulls the block forward, the other piece pulls backwards. There are no more forces. The tension is the same in a string along all its length. So you know one force. And you know the acceleration: it is 0.61 m/s2 on the left. From this you can determine the resultant force exerted by the two strings.

ehild

My T2 solution above does not include direction for acceleration. However if I include the direction (left therefore "-" 0.61m/s^2) The tension seems to small. Under the assumption that T1 > T2 > T3.

ehild
Homework Helper
At last I understood what you did. Your result T2=7.6 N is correct. And also T1>T2>T3.

The usual way to treat such problems is that we consider all bodies separately, but assume that all have the same acceleration. The direction of acceleration is along the string.

It is obvious to assume that the 1 kg mass will accelerate downward, under the effect of gravity and the tension T1 of the first string.

m1 a=m1g-T1

The second block of 2.5 kg mass is pulled to the left by the left string with force T1 and pulled to the right with the second string by force T2.

m2 a=T1-T2

The third block is pulled forward by T2 and backwards by T3

m3 a=T2-T3

The third block is pulled up by the third string with force T3 and gravity acts on it downward.

m4 a=T3-m4 g

Adding up all equations, the tensions cancel and we get that the sum of all external forces is equal to the acceleration times the sum of the masses, and this acceleration is 0.613 m/s2.

Plugging that value for a in the first equation,

T1= 1kg (9,8-0.613)=9.187 N.

T2 is obtained from the second equation:

2.5 *0.613=9.19-T2---> T2=7.654 N

and T3 comes from the third or fourth one:

4*0.613=T2-T3 --->T3=5.202 N

The fourth equation:T3-0.5 g = 0.613*0.5 --->T3=5.206 N.

The difference in the last digit comes from rounding error.

ehild