Question

In the figure we see two blocks connected by a string and tied to a wall. The mass of the lower block is 1.0 kg; the mass of the upper block is 2.0 kg; the angle of the incline is 31 degree.

Find the tension in the string connecting the two blocks.
Find the tension in the string that is tied to the wall.

Answer 1

To find the tension in the string connecting the two blocks, we consider the forces acting on each block and use trigonometry to break down the tension force. The tension in the string tied to the wall is the same as the horizontal component of the tension force in the string connecting the two blocks.

Step-by-step explanation:

To find the tension in the string connecting the two blocks, we need to consider the forces acting on each block. The lower block experiences the force of gravity pulling it downwards, and the tension force pulling it upwards. The upper block experiences the force of gravity and the tension force pulling it downwards at an angle. Using trigonometry, we can break down the tension force into vertical and horizontal components. By applying Newton's second law of motion, we can set up equations to solve for the tension in the string connecting the two blocks.

The tension in the string tied to the wall is the same as the horizontal component of the tension force in the string connecting the two blocks. Since the incline is at an angle, we can use trigonometry to calculate the horizontal component of the tension force. This component provides the necessary force to prevent the upper block from sliding down the incline.

Answer 2

1. The tension in the string connecting the two blocks
`(\(T_1\))` is approximately
`\(15.142 \, \text{N}\)`

2. The tension in the string tied to the wall
`(\(T_2\))` is approximately \
`(5.04 \, \text{N}\)`.

How to find the tension?

Given:

`\(m_1 = 1.0 \, \text{kg}\)` (mass of the lower block)

`\(m_2 = 2.0 \, \text{kg}\)` (mass of the upper block)

`\(\theta = 31 \textdegree\)` (angle of the incline)

`\(g = 9.81 \, \text{m/s}^2\)` (acceleration due to gravity)

For Block 2 (2.0 kg):

Using Newton's second law
`(\(F_{\text{net}} = 0\)` as the acceleration is zero):

`\[2 m_2 g \sin(\theta) + T_2 - T_1 = 0\]`

`\[2 m_2 g \sin(\theta) + T_2 = T_1\] ...(1)`

For Block 1 (1.0 kg):

`\[T_2 = m_1 g \sin(\theta)\] ...(2)`

Substitute equation (2) into equation (1):

`\[2 m_2 g \sin(\theta) + m_1 g \sin(\theta) = T_1\]`

`\[3 m_1 g \sin(\theta) = T_1\]`

`\[3 * 1 * 9.81 * \sin(31\textdegree) \approx 15.142 \, \text{N}\]`

Therefore, the tension in the string connecting the two blocks
`(\(T_1\))` is approximately
`\(15.142 \, \text{N}\)`.

Now, for
`\(T_2\)`:

`\[T_2 = m_1 g \sin(\theta)\]`

`\[1 * 9.81 * \sin(31\textdegree) \approx 5.04 \, \text{N}\]`

Therefore, the tension in the string tied to the wall
`(\(T_2\))` is approximately
`\(5.04 \, \text{N}\)`.