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

Answer:

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 $T_{2}8.19N$.