Atwood’s Machine. A 15.0kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0kg counterweight is suspended from the other end of the rope (Fig. E5.15). The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the tension in the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight?

The mass of bricks is $m=15\mathrm{kg}$

The mass of counterweight is $M=28\mathrm{kg}$

The acceleration of loads of bricks is calculated by considering the equilibrium of forces in a vertical direction. The tension in each rope is found using acceleration and weights of bricks and counter weight.

The expression for the force is given by,

$F=ma$

Here $m$is the mass, $F$is the force and $a$is the acceleration.

(a)

The free body diagram for bricks load is shown below,

The free body diagram for the counter weight is shown below,

(b)

The upward acceleration of the load bricks is calculated as:

$\left(m+M\right)a=\left(M-m\right)g$

Here, g is the gravitational acceleration and its value is $9.8\mathrm{m}/{\mathrm{s}}^2$.

Substitute $15\mathrm{kg}$for $m$, $28\mathrm{kg}$for $M$and $9.8\mathrm{m}/{\mathrm{s}}^2$ for g in the above equation.

$$\begin{gathered} \left(15\mathrm{kg}+28\mathrm{kg}\right)\mathrm{a}=\left(28\mathrm{kg}-15\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{a}=2.96\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the upward acceleration of the load of bricks is $2.96\mathrm{m}/{\mathrm{s}}^2$.

(c)

The tension in moving rope is calculated as:

$T=m\left(g+a\right)$

Here, g is the gravitational acceleration and its value is $9.8\mathrm{m}/{\mathrm{s}}^2$, $T$is the tension in the moving rope.

Substitute 15 kg for $m$, $9.8\mathrm{m}/{\mathrm{s}}^2$ for g and $2.96\mathrm{m}/{\mathrm{s}}^2$ for $a$in the above equation.

$$\begin{gathered} T=\left(15\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2+2.96\mathrm{m}/{\mathrm{s}}^2\right) \\ T=191\mathrm{N} \end{gathered}$$

The weight of the bricks is given as:

$$\begin{gathered} w=mg \\ w=\left(15\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{w}=147\mathrm{N} \end{gathered}$$

The weight of the load of bricks is less than the tension of moving rope.

The weight of the counterweight is given as:

$$\begin{gathered} W=Mg \\ w=\left(28\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{w}=274\mathrm{N} \end{gathered}$$

The weight of the counterweight is greater than the tension of the moving rope.

Therefore, the tension in the moving rope is 191N .