Tension in a string along with a Pseudo force

[anantchowdhary]

Homework Statement

A block of mass 50g is suspended from the ceiling of an elevator.Find the tension in the string if the elevator goes up with an acceleration of 1.2m/s^2

Homework Equations

F_{net}=ma_{net}

The Attempt at a Solution

I have reached here:...but with this i don't get the books answer...The book takes net acceleration to be 0
If F be the pseudo force...then,
T-F_g-F=ma
a=1.2
So solving this i get an answer that differs from the one given in a problem book...that is 55N(the book's ans)

Ive realized that we get 55N if we take a_{net}=0.Now how is this possible?

 

[andrevdh]

The real forces that the mass experiences are the tension in the string, T, and its weight, W. The resultant of these two forces are responsible for its given acceleration. So there is no need for a pseudo force.

Also with a mass of 50 grams the tension comes to 0.55 N. A mass of 5.0 kg will give 55 N.

 

[anantchowdhary]

The real forces that the mass experiences are the tension in the string, T, and its weight, W. The resultant of these two forces are responsible for its given acceleration. So there is no need for a pseudo force.

Also with a mass of 50 grams the tension comes to 0.55 N. A mass of 5.0 kg will give 55 N.

Well...i don't seem to get the point that u have referred to...because...we certainly need a pseudo force.But i understand that relative acceleration will be zero.

 

[HallsofIvy]

Depends upon your point of view. I would say there are two "forces" on the mass: one of mg= .05kg (9.8/s²)= 0.49 N downward and the "pseudo" force of ma= 0.5kg(1.2 m/s<sup>2)= 0.06 N, also downward, so that the total "force" on the spring and its tension is 0.06+ 0.49= 0.55 N.

Of course, you could also have said that there is a net "acceleration" of 11 m/s2 downward given F= ma= 0.55N.</sup>

 

[anantchowdhary]

Thnx for the solution to the problem