Physics Pulley homework

  • Thread starter crays
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  • #1
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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?
 

Answers and Replies

  • #2
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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)
 
  • #3
tiny-tim
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Homework Helper
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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:
 
  • #4
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oops. looks like i read the data wrong!
sorry 'bout that :)
 
  • #5
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Thanks !! I've found the answer. But i don't understand why the tension 1500 is distributed to all over the string.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
251
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:
 
  • #7
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Ah, thanks. Much more understandable.
 

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