# What is the tension in the lift cable and in the string?

• Starrrrr
In summary, the tension in the lift cable and in the string is 324 N. If the lift accelerates upwards at 1 m/s2, the tension in the lift cable is at least 3240 N so that the string remains intact.
Starrrrr
1.
Part One:

A mass M1 = 30 kg is suspended by a light string from the ceiling of a lift of mass M2 = 300kg.
If the lift accelerates upwards at 1 m/s2, what is the tension in the lift cable and in the string? (g=9.8 m/s)

Part Two:
If the string breaks when subjected to a tension of more than 700 N, what is the greatest possible tension in the lift cable so that the string remains intact?

2. Homework Equations : t-mg=ma , f=ma3. The attempt at a solution: Part One: Tension in string, t-mg=ma, t-(30)(9.8)=30(1), t=324N
Tension in lift cable, t-mg=ma, t-(300)(9.8)=300(1), t=3240N ( Is this correct ?)

Part two: Not sure about part two.

Doesn't the lift cable have to support the weight of the 30 kg mass also?

TomHart said:
Doesn't the lift cable have to support the weight of the 30 kg mass also?
Your right! So the combined mass is 330kg,then the tension= 3564N and the second tension force equals 3240N ?.

Starrrrr said:
the second tension force equals 3240N ?.
Can you please explain what you mean by "second tension force"?

TomHart said:
Can you please explain what you mean by "second tension force"?
Lol, sorry I mean the tension of the string

It looks like your result for tension of the string is off by a factor of ten. Can you take another look at that. Maybe it's just a typo.

TomHart said:
It looks like your result for tension of the string is off by a factor of ten. Can you take another look at that. Maybe it's just a typo.
How? I calculated the tension of string=(300)(1+9.8)=3240N

What mass is the string supporting? 30 kg or 300 kg?

Starrrrr
TomHart said:
What mass is the string supporting? 30 kg or 300 kg?
Oh the mass of the string its 30 kg. 324N. Sorry, for the confusion.

TomHart
Starrrrr said:
Oh the mass of the string its 30 kg. 324N. Sorry, for the confusion.
For the second part, calculate the minimum acceleration of the lift for which the string will break.

Starrrrr
Buffu said:
For the second part, calculate the minimum acceleration of the lift for which the string will break.
so a=f/m = 700/330=2.12ms^-2

I thought the string only supported the 30 kg mass.

TomHart said:
I thought the string only supported the 30 kg mass.
Oh I thought when buffu said 'acceleration of the lift for which the string will break' I had to combine the masses, I interpreted it wrong.

Also consider the directions of your accelerations. Which way does g point? What's the direction of the mass' acceleration as it's being pulled by the cable?

VivaLaFisica said:
Also consider the directions of your accelerations. Which way does g point? What's the direction of the mass' acceleration as it's being pulled by the cable?
ok thanks.

Last edited:
so is the accerelation equal to=23.33m^-2

Starrrrr said:
t-mg=ma
This is the equation you originally used. You should use this same equation for this calculation. I'm sorry I didn't point this out sooner, but I just now noticed it.

TomHart said:
This is the equation you originally used. You should use this same equation for this calculation. I'm sorry I didn't point this out sooner, but I just now noticed it.
Thats fine, thank you for your help :)

Just to explain a little bit, the answer you calculated (23.33 ms-2) would be the right answer for a mass of 30 kg sitting on a frictionless surface with the 700 N force acting horizontally. In the situation of this problem, the weight of the 30 kg mass is acting against the tension of the string so the acceleration is reduced accordingly. Contrast that with a problem where the 30 kg mass had a 700 N tension acting on it downward (in the direction of gravity), In that case, the 700 N force would produce a greater acceleration than the other two cases.

Starrrrr
Starrrrr said:
Thats fine, thank you for your help :)
Why not make a FBD for these kinds of problem ?

## 1. What is tension in relation to lift cables and strings?

Tension refers to the force exerted on an object by a pulling or stretching motion. In the case of lift cables and strings, tension is the force that is applied to these objects to support the weight of the lift or whatever is being lifted.

## 2. How is tension measured in lift cables and strings?

Tension is typically measured in units of force, such as pounds or Newtons. In lift cables and strings, tension can be measured using a tension meter or by calculating the force using the equation F = m x a (force = mass x acceleration).

## 3. What factors affect the tension in lift cables and strings?

The tension in lift cables and strings can be affected by various factors such as the weight of the object being lifted, the angle of the cable or string, and the strength and elasticity of the material used for the cable or string.

## 4. How does tension impact the safety of lift cables and strings?

Tension is a critical factor in the safety of lift cables and strings. If the tension is too low, the lift may not be able to support the weight of the object, leading to potential accidents. On the other hand, if the tension is too high, it could cause the cables or strings to break, also resulting in potential hazards.

## 5. How can tension in lift cables and strings be controlled?

Tension in lift cables and strings can be controlled by adjusting the weight of the object being lifted, using stronger and more durable materials for the cables and strings, and ensuring proper maintenance and regular inspections of the lift system.

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