Question

8. A 300g mass hangs at the end of a string. A second-string hang from the bottom of that mass and supports a 900g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700m/s2 . (b) Find the tension in each string when the masses are accelerating downward at 0.700m/s2 .

Answer 1

We are looking for:
Tension A=?
Tension B=?
Then convert grams to kilograms
1kg = 1000g
300g= 0.3 kg
900g= 0.9kg
Given:
mA= 0.3kg
mB= 0.9kg
how? W=mg
g= 9.8 m/s^2
ay= 0.700m/s^2
WA= 2.94N

WB= 8.82N
For A [Tension A-Weight A - Tension B=mAay]
TA - 2.94 - TB=0.3kg(0.700m/s^2)
TA - 2.94 - TB = 0.21 N
TA - TB = 0.21N + 2.94 N
TA - TB = 3.15 N
* refer TB from below*
TA - TB = 3.15N
TA - 9.45N = 3.15N
TA = 3.15N + 9.45N
TA = 12.6N

For B F = ma
Summation of forces along y-axis = mBay
-8.82 N + TB = 0.9 kg(0.700 m/s^2)
-8.82N + TB = 0.63 N
TB = 0.63 N + 8.82 N
TB = 9.45 N

Answer 2

First we need to analyse each mass and all the forces that interact with each of them. To do that, we use a free-body diagram and from that it'll be easier to analyse all the forces and write the mathematical relation between them, each free-body diagram is attached.

From these diagrams and using Newton's Law
`\sum F=ma`, we define a equation for each mass.

Mass 1:

`\sum F_{y_(1)}= m_(1) a`

`T_(1)-T_(2)-W_(1)=m_(1) a\\T_(1)=m_(1) a+T_(2)+W_(1)\\T_(1)=(0.3kg)(0.7m/s^(2) )+T_(2)+(0.3kg)(9.8m/s^(2) )\\T_(1)=(0.21 +T_(2)+2.94)N\\T_(1)=(3.15+T_(2))N`

Now, we need to find a expression for the second tension and solve the equation.

Mass 2:

`\sum F_{y_(2) }= m_(2) a\\T_(2)-W_(2) =m_(2) a\\T_(2)=m_(2) a+W_(2)\\T_(2)=(0.90kg)(0.7m/s^(2))+(0.90kg)(9.8m/s^(2) )\\T_(2)=(0.63+8.82)N\\T_(2)=9.45N`

Now, we replace the second tension in the first equation:

`T_(1)=(3.15+T_(2))N=(3.15+9.45)N=12.6N`

Therefore, with an upward acceleration, the tensions are
`T_(2)=9.45N` and
`T_(1)=12.6N`.

On the other hand, we downwards acceleration, we just need to change 0.7 by -0.7, and then solve.

`T_(2)=(0.90kg)(-0.7m/s^(2))+(0.90kg)(9.8m/s^(2) )=(-0.63+8.82)N=8.19N`.

`T_(1)=(0.3kg)(-0.7m/s^(2) )+T_(2)+(0.3kg)(9.8m/s^(2))=(-0.21+8.19+2.94)N=10.92`.

Therefore, with downwards acceleration, tensions would be
`T_(1)=10.92` and
`T_(2)8.19N`.