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

[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.

 

[TomHart]

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

 

[Starrrrr]

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 ?.

 

[TomHart]

the second tension force equals 3240N ?.

Can you please explain what you mean by "second tension force"?

 

[Starrrrr]

Can you please explain what you mean by "second tension force"?

Lol, sorry I mean the tension of the string

 

[TomHart]

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.

 

[Starrrrr]

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

 

[TomHart]

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

 

[Starrrrr]

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.

 

[Buffu]

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]

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

 

[TomHart]

I thought the string only supported the 30 kg mass.

 

[Starrrrr]

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.

 

[VivaLaFisica]

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?

 

[Starrrrr]

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.

 

[Starrrrr]

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

 

[TomHart]

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.

 

[Starrrrr]

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 :)

 

[TomHart]

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.

 

[Buffu]

Thats fine, thank you for your help :)

Why not make a FBD for these kinds of problem ?