# Homework Help: Dynamics pully confusion

1. Dec 30, 2011

### ja!mee

Okay So, the question I have here is..

Masses are connected as shown. m1 = 5.0 kg and m2 = 4.0 kg.
a) find the acceleration of each object and
b) the tension in the two strings

I have attached the picture example as well (as there is another unlisted weight in it) With pulley problem's I find it easier to find the acceleration first, so I have done so with:

a = Fg7- Fg5/Mt
a = (7kg)(9.8 m/s2) - (5kg)(9.8 m/s2) / 12 kg
where Fg7 is the force of gravity on the 4kg an 3 kg weigh combined and Fg5 is the force of gravity with the 5 kg weight.
I have found the the correct answer for question A) which is acceleration = 1.63

Now with that said an done I am not looking for the rope tension, I have my equation:

T = Fg7 - ma
T = 7 kg (9.8 + 1.6)
T = 79.8 N

Well that's all fine and dandy, but I am in a course where the answers are provided, it is the work that you do to get to the answer that really counts. The answer I have provided is not 79.8 N but 57N.

The only way I could see getting to this answer is
T = Fg5 - ma
T = 5 kg (9.8 + 1.6)
T = 57 N

It is my understanding that the tension equation that I take should come from the heaviest set of weights on any given side of the pulley. From the picture provided, you can see that there are two sets of weights on the one side of the pulley. Am I wrong in thinking that to find the tension I should add those two weights together and find the Tension from that number? Because this seems to contradict what the answer I have been provided is saying.

Or am I missing something completely?

#### Attached Files:

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2. Dec 30, 2011

### Staff: Mentor

In the second step you added when you should have subtracted.

Better to do things systematically:
ƩF = ma
Taking up as positive, the acceleration is -1.63 m/s2:
T - mg = m(-1.63), and so on.

Systematically:
ƩF = ma
Taking up as positive, the acceleration is +1.63 m/s2:
T - mg = m(1.63), and so on.

You can use either side to find the tension, since it's the same. You just made an error with signs.

3. Dec 30, 2011

### technician

Lookat m1 (5kg). You have its acceleration upwards =1.63. The forces acting on m1 are the upward force due to tension =T and the downward force due to its weight = 5 x 9.81
therefore for m1 resultant force (upward) T - 5x9.81 = 5 x 1.63
this gives T = 57.2N
For the string between m2 and the 3kg mass you need to look at the 3kg mass.
The 3kg mass is accel;erating down at 1.63 therefore the resultant force on the 3kg is (3 x 9.81) - tension
Therefore 29.43 - T = 3 x 1.63
gives T = 24.54N

4. Dec 30, 2011

### ja!mee

Thank you!!! that absolutely clears things up.

5. Dec 30, 2011

### technician

You can check these numbers by looking at m2 (4kg)
It is accelerating down with acc = 1.63
The forces acting on m2 are the tension in the main string (57.2N upwards) with the weight (4x9.81) plus the tension in the other string (24.54) downwards.
therefore the resultant force downwards = 39.24 + 24.54 -57.2 = 6.58N
F = ma 6.58 = 4 x 1.63 (gives 6.52....sig figs)

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