Newton law

Oct 2019
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I have a question and I need help. It is as follows:

A long rope of 15 meters hangs on a tree. A 50 kg mass can hang on the rope; any additional mass will cause the rope to break. How long will it take you to pull a 30 kg mass box up the tree.

What formula can be used to calculate this?
 

topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
I have a question and I need help. It is as follows:

A long rope of 15 meters hangs on a tree. A 50 kg mass can hang on the rope; any additional mass will cause the rope to break. How long will it take you to pull a 30 kg mass box up the tree.

What formula can be used to calculate this?
This is not a well stated problem. We can haul the mass at any constant speed and the rope will be fine. I'm going to assume you want the mass lifted as quickly as we can and I'm going to assume some positive upward acceleration a.

So. The rope can handle 50g of tension and the given mass puts 30g on it. That means we can deal with an acceleration of up to a, where 50g = 30g + 30a.

Can you finish?

-Dan
 
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Oct 2019
3
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It means I have to work out like terms and isolate the a. With the a (m/s), then I work out how many seconds it will take for the 15 meters.

Am I on the right path?
 
Jun 2016
1,142
514
England
Yes but remember "a" is an acceleration (m/s/s)

Topsquark was working out the maximum force that could be applied before the rope breaks.

Force = Mass * Acceleration

Maximum Force = 50kg * acceleration due to Gravity

Box weighs 30 kg so we have some force left over before the rope breaks.

Maximum Force = 30 kg * acceleration due to gravity + 30kg * acceleration due to pulling on the rope.
 
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topsquark

Forum Staff
Apr 2008
2,926
608
On the dance floor, baby!
It means I have to work out like terms and isolate the a. With the a (m/s), then I work out how many seconds it will take for the 15 meters.

Am I on the right path?
Yes. Once you get the acceleration then you can use \(\displaystyle y = y_0 + v_0 t + (1/2)at^2\) (or s = ut + (1/2)at^2 or whatever form you've been taught.)

-Dan