Calculating Tension in a Pulley System

AI Thread Summary
In the discussion about calculating tension in a pulley system, a worker of 80kg pulls a 40kg platform upward with an acceleration of 2.5 m/s². The initial calculations confused the forces acting on the system, leading to incorrect assumptions about mass and tension. It was clarified that the man and platform should be treated as a single object, with tension being the same throughout the string due to its properties. The relationship between the force exerted by the worker and the resulting movement of the platform was explained, emphasizing the conservation of energy. Ultimately, the correct understanding of tension distribution and forces resolved the initial confusion.
crays
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Hello guys, i have a question i would like to ask.

A worker of mass 80kg stands in a platform of mass 40kg pulls the platform up with an acceleration of 2.5ms^-2 using a smooth pulley as shown in the figure. what is the tension T in each side of the rope (g = 10 ms^-2).

The figure : A man standing in a platform, the platform is attached to a string which is over a pulley and the man is pulling the other side of the string.

I thought of F-mg = ma
which will be
F = (120)(2.5) + 1200

but its wrong @_@.

Can someone tell me?
 
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from what i understand you took m=80+40.
but the man is not accerlerating, only the platform.
so m should be 40.
so the equations should look like this:
1) F=ma
2) F=T-mg
from here it's easy:
ma=T-mg
40*2.5=T-10*40
T=40*12.5
notice that i used a capital T to denote the tention in the rope. i used the capital F to denote the total force acting on the platform.

F=ma is Newton's law: ma=total force. this is the reason for eq. (1)
but the total force is the rope pulling up (T) and gravity pulling down(-mg). this is eq. (2)
 
crays said:
A worker of mass 80kg stands in a platform of mass 40kg pulls the platform up with an acceleration of 2.5ms^-2 using a smooth pulley as shown in the figure. what is the tension T in each side of the rope (g = 10 ms^-2).

I thought of F-mg = ma
which will be
F = (120)(2.5) + 1200

Hi crays! :smile:

Yes, your'e right, the man is accelerating with the platform.

And your F is correct.

But T is not equal to F.

You haven't drawn a proper diagram for yourself, have you?

Hint: treating the man-and-platform as one object, how many forces upward are there? :wink:
 
oops. looks like i read the data wrong!
sorry 'bout that :)
 
Thanks ! I've found the answer. But i don't understand why the tension 1500 is distributed to all over the string.
 
crays said:
Thanks ! I've found the answer. But i don't understand why the tension 1500 is distributed to all over the string.

Hi crays! :smile:

Just look at your diagram:

it should show that both ends of the string are attached to the man-platform body, both pulling vertically upwards;

the tension T is the same throughout the string ('cos that's the way strings work),

so there's a force T upward at both ends of the string, and both those forces are acting on the body. :smile:

(usually, the other end of the string is attached to something external, like a ceiling, so the force there isn't acting on the body at all)

Another way of looking at this is that there is gearing: when the man pulls the rope one foot, the platform only moves half a foot doesn't it?

How come the man exerts a force of T, but the platform moves under a force of 2T … where does the extra energy come from? :confused:

Because his force T does T foot-pounds of work, while the work done on the platform is 2T times half a foot-pound, which is the same … no extra energy! :smile:
 
Ah, thanks. Much more understandable.
 

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