System of Masses and Pulley, Problem

  • Thread starter fsun
  • Start date
  • #1
2
0

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
    System of Masses Problem.JPG
    5.3 KB · Views: 383
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,914
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
 
  • #3
2
0
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.
 
  • #4
ehild
Homework Helper
15,543
1,914
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
 

Related Threads on System of Masses and Pulley, Problem

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
5K
Replies
2
Views
276
  • Last Post
Replies
14
Views
16K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
15
Views
926
Replies
6
Views
2K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
4
Views
1K
Top