Question

If you hold a 3.89-kg package by a light vertical string, what will be the tension in this string when the elevator accelerates as in the previous part?

Answer 1

a) 784.5 N

b) 47.8 N

Explanation:

a)

Find the free-body diagram of the person in attachment.

There are two forces acting on the person:

- Its weight (gravitational force), of magnitude W = 625 N, downward

- The normal reaction exerted by the scale on the person, upward, of magnitude N: this corresponds to the reading on the scale

Therefore, the net upward force on the person is

$F=N-W$

According to Newton's second law of motion, the net force must be equal to the product of mass and acceleration, so:

$F=N-W=ma$

where:

$m=\frac{W}{g}=\frac{625}{9.8}=63.8 kg$ is the mass of the person

$a=2.50 m/s^2$ is the acceleration of the elevator (upward)

Re-arranging the equation and solving for N, we can find the reading on the scale (the normal force):

$N=ma+W=(63.8)(2.50)+625=784.5 N$

b)

In this case, you are holding a package of 3.85 kg with a light string.

The forces acting on the package are:

- The weight (force of gravity), $W=mg$, downward

- The tension in the string, T, upward

As before, the net force on the package is

$F=T-W=T-mg$

And according to Newton's second law, it must be equal to the product of mass and acceleration:

$T-mg=ma$

Therefore here we have:

m = 3.89 kg is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity

$a=2.50 m/s^2$ is the acceleration of the elevator

Solving for T, we find the tension in the string:

$T=m(a+g)=(3.89)(2.50+9.8)=47.8 N$