Newtons Laws EX 18

(a) the blocks are at rest, that is $a=0$.

To find the value of $F_{T_{1}}$, we look at the two forces, $F_{T_{1}}$ and $F_{G}$, acting on the system comprised of both masses, $m_{A}$ and $m_{B}$, as shown:

$F_{T_{1}}-5N=0$

$F_{T_{1}}=5N$

To find the value of $F_{T_{2}}$, we isolate $m_{B}$:

$F_{T_{2}}-3N=0$

$F_{T_{2}}=3N$

(b) the blocks are moving at constant velocity, that is $a=0$!. $F_{T_{1}}=5N$ and $F_{T_{2}}=3N$ as above.

(c) The blocks accelerate upwards at $2m/s^{2}$, that is, they have an acceleration of $+2m/s^{2}$.

To find the value of $F_{T_{1}}$, we look at the forces acting at the system comprised of both masses, $m_{A}$ and $m_{B}$, as shown in part a):

$F_{T_{1}}-5N=(m_{A}+m_{B})a$

$F_{T_{1}}-5N=(0.5kg)(2m/s^{2})$

$F_{T_{1}}=5N+1N=6N$

To find the value of $F_{T_{2}}$, we isolate $m_{B}$ and look at the forces acting on it:

$F_{T_{2}}-3N=m_{B}a$

$F_{T_{2}}-3N=(0.3kg)(2m/s^{2})$

$F_{T_{2}}=3N+0.6N=3.6N$

(d) The blocks accelerate downwards at $2m/s^{2}$, that is they have an acceleration of $-2m/s^{2}$. Similarly to part c), we can find the value of $F_{T_{1}}$ by looking at the forces acting at the system comprised of both masses, $m_{A}$ and $m_{B}$:

$F_{T_{1}}-5N=(m_{A}+m_{B})a$

$F_{T_{1}}-5N=(0.5kg)(-2m/s^{2})$

$F_{T_{1}}=5N-1N=4N$

To find the value of $F_{T_{2}}$, we isolate $m_{B}$ and look at the forces acting on it:

$F_{T_{2}}-3N=m_{B}a$

$F_{T_{2}}-3N=(0.3kg)(-2m/s^{2})$

$F_{T_{2}}=3N-0.6N=2.4N$