Tension of a rope in a pulley system

In summary: The heavy mass' equation of force will be:-Mg+T=-Ma-(4 kg)(9.8 m/s^2)+T=-(4kg)(3.60 m/s^2)that's it, u've found it :)
  • #1
winthos
2
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)?
 

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  • #2
can u draw the force diagrams of each mass? that will help u a lot :)
 
  • #3
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:
  • #4
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? :)
 

1. What is a pulley system?

A pulley system is a mechanical device that uses a grooved wheel and a rope or cable to lift or move heavy objects. It is commonly used in cranes, elevators, and other lifting devices.

2. How does a pulley system work?

A pulley system works by distributing the weight of an object across multiple ropes or cables, making it easier to lift or move. As one end of the rope is pulled, the object is lifted by the other end of the rope through the pulley.

3. What is tension in a pulley system?

Tension in a pulley system refers to the force that is exerted on the rope or cable as a result of the weight of the object being lifted. It is the force that keeps the rope taut and allows the system to function properly.

4. How is tension calculated in a pulley system?

The tension in a pulley system can be calculated using the formula T = W / n, where T is the tension, W is the weight of the object being lifted, and n is the number of ropes or cables supporting the object. This formula assumes that the pulleys are frictionless and the ropes are weightless.

5. What factors affect the tension in a pulley system?

The tension in a pulley system is affected by several factors, including the weight of the object being lifted, the number of ropes or cables supporting the object, the angle of the ropes, and the friction between the ropes and the pulleys. These factors can impact the efficiency and safety of the pulley system.

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