Tension of a rope in a pulley system

AI Thread Summary
In a pulley system involving a 1.0 kg and a 4.0 kg mass, the acceleration of the larger mass was calculated to be 5.88 m/s² using the net force equation. The tension in the cord is derived from the forces acting on the masses, leading to confusion about which mass to use for calculations. It was clarified that either mass can be used to find tension, but focusing on the heavier mass simplifies the process. The equation for the heavier mass shows how to calculate tension based on its weight and acceleration. Ultimately, both masses can be utilized to arrive at the correct tension value.
winthos
Messages
2
Reaction score
0

Homework Statement


A 1.0 kg mass and a 4.0 kg mass are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. What is the acceleration of the larger mass? What is the tension on the cord?

Homework Equations



F(net force)=F(tension)-weight

F=ma


The Attempt at a Solution


I found the acceleration using F=ma. The system has 3 kg of mass going down (4kg-1kg), which is equal to 29.4N of force (found by multiplying 3kg by 9.8m/s^2). The whole system has a mass of 5 kg, so I substituted into F=ma.

29.4N=5.0 kg + a

The acceleration is 5.88 m/s^2

Now comes the part I am confused about. The tension of the cord should be
F(net force)=tension-weight, or rearranged to be:

F(tension)=F(net force)+weight or (Tension=ma+mg)

I can substitute in what i know:
Tension =m(5.88m/s^2)+m(9.8m/s^2)

I am at a loss of which mass to use. Do i use the mass that i found in working on acceleration (3 kg), or the mass of the whole system (5 kg)?
 

Attachments

  • forcediag.jpg
    forcediag.jpg
    10.9 KB · Views: 3,287
Last edited:
Physics news on Phys.org
can u draw the force diagrams of each mass? that will help u a lot :)
 
I attached a quick diagram, i don't know if that's detailed enough, but that is basically what's going on. (i'm bad at drawing with a mouse, so it probably doesn't look that spectacular)- the 4kg mass is pulling the whole system down at 5.88m/s^2, which is great since that is the first half of the question.

My problem is which mass do i use to find the tension? the mass of 3 kg (because of the pulley 4kg-1kg=3kg, the 3kg was then used to find the amount of downward force, which turned out to be 29.4N) or the mass of the whole system, 5kg?
 
Last edited:
winthos said:
I attached a quick diagram, i don't know if that's detailed enough, but that is basically what's going on. (i'm bad at drawing with a mouse, so it probably doesn't look that spectacular)- the 4kg mass is pulling the whole system down at 5.88m/s^2, which is great since that is the first half of the question.

My problem is which mass do i use to find the tension? the mass of 3 kg (because of the pulley 4kg-1kg=3kg, the 3kg was then used to find the amount of downward force, which turned out to be 29.4N) or the mass of the whole system, 5kg?

ok, u are done basically, u can use any of the masses actually, so let's say we concentrate on the heavier one

the heavy mass' equation of force will be:

-Mg+T=-Ma

-(4 kg)(9.8 m/s^2)+T=-(4kg)(5.88 m/S^2)

that's it, u've found it :)

can u find it now using the smaller mass? :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
13
Views
2K
Replies
3
Views
2K
Replies
38
Views
4K
Replies
18
Views
4K
Replies
27
Views
2K
Replies
30
Views
3K
Replies
18
Views
5K
Replies
22
Views
6K
Back
Top