Question:Two blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes (Fig. P5.90). (a) Which way will the system move when the blocks are released from rest? (b) What is the acceleration of the blocks? (c) What is the tension in the cord?

The given data can be listed below as:

• The mass of the bigger block is, $M=100\text{kg}$.
• The mass of the smaller block is, $m=50\text{kg}$.
• The angle made by the bigger block with the horizontal is, ${\theta }_2=30.0°$.
• The angle made by the smaller block with the horizontal is, ${\theta }_1=53.1°$.

The acceleration is described as the change of the velocity with respect to the time. The acceleration is described as the division of the force exerted and the mass of that object.

Let us consider that the system will move to the left.

The free body diagram of the system has been drawn below:

From the above diagram, it can be the summation of the forces acting on both the masses is equal to zero in each direction.

The equation of the summation of the forces acting on the bigger block is expressed as:

$$\begin{gathered} Mg\sin {\theta }_2-T=Ma \\ T=Mg\sin {\theta }_2-Ma.....\left(i\right) \end{gathered}$$

Here, T is the tension acting on the blocks, m is the mass of the bigger block, ${\theta }_2$is the angle made by the smaller block with the horizontal and a is the acceleration of the blocks.

The equation of the summation of the forces acting on the smaller block is expressed as:

$$\begin{gathered} T-mg\sin {\theta }_1=ma \\ T=mg\sin {\theta }_1+ma.....\left(ii\right) \end{gathered}$$

Here, T is the tension acting on the blocks, m is the mass of the smaller block, ${\theta }_1$is the angle made by the bigger block with the horizontal and a is the acceleration of the blocks.

Equating the equation (i) and (ii) to get the value of the acceleration on the blocks

$$\begin{gathered} Mg\sin {\theta }_2-Ma=mg\sin {\theta }_1+ma \\ a\left(m+M\right)=Mg\sin {\theta }_2-mg\sin {\theta }_1 \\ a=\frac{g\left(M\sin {\theta }_2-m\sin {\theta }_1\right)}{m+M}\dots ..\left(iii\right) \end{gathered}$$

From the above equation, it is evident that the acceleration is positive. Hence, the system will move in the left direction.

Thus, the system will move in the left direction when the blocks are released from rest.

The equation (iii) is recalled below:

$a=\frac{g\left(M\sin {\theta }_2-m\sin {\theta }_1\right)}{m+M}$

Substitute all the values in the above equation.

$\begin{array}{rcl}a& =& \frac{\left(9.8{\text{m/s}}^{\text{2}}\right)\left[\left(100\text{kg}\right)\sin 30.0°-\left(50\text{kg}\right)\sin 53.1°\right]}{50\text{kg}+100\text{kg}}\\ & =& \frac{\left(9.8{\text{m/s}}^{\text{2}}\right)\left[\left(100\text{kg}\right)\left(0.5\right)-\left(50\text{kg}\right)\left(0.79\right)\right]}{150\text{kg}}\\ & =& \frac{\left(9.8{\text{m/s}}^{\text{2}}\right)\left[\left(50\text{kg}\right)-\left(39.5\text{kg}\right)\right]}{150\text{kg}}\\ & =& \frac{\left(9.8{\text{m/s}}^{\text{2}}\right)\left(10.5\text{kg}\right)}{150\text{kg}}\end{array}$

Hence, further as:

$\begin{array}{rcl}a& =& \frac{\left(9.8{\text{m/s}}^{\text{2}}\right)\left(10.5\text{kg}\right)}{150\text{kg}}\\ & =& \left(9.8{\text{m/s}}^{\text{2}}\right)\left(0.07\right)\\ & =& 0.686{\text{m/s}}^{\text{2}}\end{array}$

Thus, the acceleration of the blocks is $0.686{\text{m/s}}^{\text{2}}$.

Substitute$3.528{\text{m/s}}^{\text{2}}$ for in the above equation.

$\begin{array}{rcl}T& =& \left(100\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\sin 53.1°-\left(100\text{kg}\right)\left(3.528{\text{m/s}}^{\text{2}}\right)\\ & =& \left(980\text{kg}·{\text{m/s}}^{\text{2}}\right)\left(0.79\right)-352.8\text{kg}·{\text{m/s}}^{\text{2}}\\ & =& \left(774.2\text{kg}·{\text{m/s}}^{\text{2}}\right)-352.8\text{kg}·{\text{m/s}}^{\text{2}}\\ & =& 421.4\text{kg}·{\text{m/s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 421.4\text{J}\end{array}$

Thus, the tension in the cord is $421.4\text{J}$.