Question:

A block of mass $m$ is suspended by two ropes, as shown in the figure below. The angle formed between each rope and the horizontal axis is $\theta$. The block is in equilibrium.

(a) Derive an expression for the tension $T_1$ in terms of the given quantities.

(b) Derive an expression for the normal force $N$ in terms of the given quantities.

(c) Calculate the tension $T_1$ and the normal force $N$ when $\theta = 30^\circ$, $m = 2 \, \text{kg}$, and the acceleration due to gravity $g = 9.8 \, \text{m/s}^2$.

Explanation:

(a) To derive an expression for the tension $T_1$, let's consider the forces acting on the block in the vertical direction. These forces are the tension $T_1$, the tension $T_2$, and the weight $mg$.

The net force in the vertical direction is $0$ since the block is in equilibrium. Therefore, the sum of the vertical forces must be zero:

$T_2 \cos(\theta) - T_1 \cos(\theta) - mg = 0$

Simplifying this equation, we can express $T_1$ in terms of the given quantities:

$T_1 = T_2 - mg$

Since there is no horizontal acceleration, $T_2$ can be expressed as:

$T_2 = T_1 \sin(\theta)$

Substituting this expression for $T_2$ into the previous equation, we get:

$T_1 = T_1 \sin(\theta) - mg$

Simplifying, we can solve for $T_1$:

$T_1(1 - \sin(\theta)) = mg$

Dividing both sides by $(1 - \sin(\theta))$, we obtain the expression for $T_1$:

$T_1 = \frac{mg}{1 - \sin(\theta)}$

(b) To derive an expression for the normal force $N$, let's consider the forces acting on the block in the horizontal direction. The only horizontal force is the tension $T_2$.

The net force in the horizontal direction is $0$ since the block is in equilibrium. Therefore, the sum of the horizontal forces must be zero:

$T_2 \sin(\theta) = N$

Thus, we can express the normal force $N$ in terms of the given quantities as:

$N = T_2 \sin(\theta)$

(c) To calculate the tension $T_1$ and the normal force $N$, we substitute the given values into the derived equations.

Given: $\theta = 30^\circ$, $m = 2 \, \text{kg}$, and $g = 9.8 \, \text{m/s}^2$.

From part (a), the expression for $T_1$ is:

$T_1 = \frac{mg}{1 - \sin(\theta)}$

$T_1 = \frac{(2 \, \text{kg})(9.8 \, \text{m/s}^2)}{1 - \sin(30^\circ)}$

$T_1 = \frac{19.6 \, \text{N}}{1 - 0.5}$

$T_1 = 39.2 \, \text{N}$

From part (b), the expression for $N$ is:

$N = T_2 \sin(\theta)$

$N = T_1 \sin(30^\circ)$

$N = 39.2 \, \text{N} \cdot 0.5$

$N = 19.6 \, \text{N}$

Therefore, when $\theta = 30^\circ$, $m = 2 \, \text{kg}$, and $g = 9.8 \, \text{m/s}^2$, the tension $T_1$ is $39.2 \, \text{N}$ and the normal force $N$ is $19.6 \, \text{N}$.