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:

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

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

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

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

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

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

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

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):

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

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

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

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

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

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

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

(d) The blocks accelerate downwards at $2 m / s^{2}$ , that is they have an acceleration of $- 2 m / s^{2}$ . 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 $m_{B}$ :

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

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

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

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

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

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

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

Newtons Laws EX 18

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