The ends of a light string are tied to two hooks A and B in a ceiling which are 100 cm apart horizontally so that the length between hooks is 140 cm. A 650g mass is then attached by a second length of string to a point C on the first, 80 cm from A, and hangs freely. Find, by drawing or calculation, the tensions in the portions AC and BC of the string.

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To find the tensions in the portions AC and BC of the string, we can consider the forces acting on the mass at point C.

Let's start by drawing a diagram to visualize the situation:

```
A --------- C --------------------- B
\___________/
80 cm
```

Now, let's consider the forces acting on the mass:

1. Tension in the AC portion of the string, let's call it T₁.
2. Tension in the BC portion of the string, let's call it T₂.
3. Weight of the mass (650g) acting downward, which can be calculated as: W = m * g, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).

Since the mass is hanging freely, the net force in the vertical direction must be zero. Therefore, the sum of the vertical component of the tensions and the weight must be zero.

Now, let's calculate the tensions using these considerations:

1. Tension in AC (T₁):
The string AC forms a right-angled triangle with the vertical line. The vertical component of T₁ balances the weight of the mass.

Using trigonometry, we can find the vertical component of T₁:
cosθ = adjacent / hypotenuse
cosθ = 80 cm / 140 cm
cosθ ≈ 0.5714

The vertical component of T₁ can be calculated as:
T₁(vertical) = T₁ * cosθ

2. Tension in BC (T₂):
The string BC also forms a right-angled triangle with the vertical line. The vertical component of T₂ will be equal to the vertical component of T₁, as the weight of the mass is balanced by the net vertical component of the tensions.

Hence, T₂(vertical) = T₁(vertical)

Now, let's calculate the tensions:

1. Tension in AC (T₁):
We already have the vertical component of T₁, so we can find T₁ using trigonometry:
sinθ = opposite / hypotenuse
sinθ = 80 cm / 140 cm
sinθ ≈ 0.5714

The horizontal component of T₁ can be calculated as:
T₁(horizontal) = T₁ * sinθ

2. Tension in BC (T₂):
Since T₂(vertical) = T₁(vertical), we can directly use the vertical component of T₁ to find T₂(vertical).

Finally, we can calculate the tensions:
T₁ = sqrt(T₁(horizontal)² + T₁(vertical)²)
T₂ = sqrt(T₂(horizontal)² + T₂(vertical)²)

Let's calculate the values step-by-step:

1. Calculate the vertical component of T₁:
T₁(vertical) = 650g = 0.65 kg * 9.8 m/s² = 6.37 N

2. Calculate the horizontal component of T₁:
T₁(horizontal) = T₁(vertical) * tanθ
T₁(horizontal) = 6.37 N * (80 cm / 140 cm)
T₁(horizontal) ≈ 3.656 N

3. Tension in AC (T₁):
T₁ = sqrt(T₁(horizontal)² + T₁(vertical)²)
T₁ = sqrt((3.656 N)² + (6.37 N)²)
T₁ ≈ 7.449 N

4. Tension in BC (T₂):
T₂ = T₁
T₂ ≈ 7.449 N

Therefore, the tension in the AC portion of the string is approximately 7.449 N, and the tension in the BC portion of the string is also approximately 7.449 N.

To find the tensions in the portions AC and BC of the string, we can use the principles of equilibrium.

1. Start by drawing a diagram of the situation. Label point A as the left hook, point B as the right hook, and point C as the point where the mass is attached. Draw the horizontal distance between A and B as 100 cm and the vertical distance from A to C as 80 cm.

2. Divide the string into two sections: AC and BC. Let's assume the tension in AC is T1 and the tension in BC is T2.

3. Consider the forces acting on point C. There are two forces: the tension T1 acting vertically upward and the weight of the 650g mass acting vertically downward. Since the mass is in equilibrium, the vertical forces must balance each other.

Therefore, T1 = Weight of the mass.

4. Calculate the weight of the mass. The weight is given by the formula weight = mass × acceleration due to gravity. Here, the mass is 650g, so we need to convert it to kilograms by dividing it by 1000. The acceleration due to gravity is approximately 9.8 m/s^2.

Weight = (650g / 1000) kg × 9.8 m/s^2

5. Calculate the tension in the portion AC (T1). Since T1 is equal to the weight of the mass and is acting vertically upward, the tension in AC is also acting vertically downward. Therefore, the tension in AC can be calculated by:

T1 = (650g / 1000) kg × 9.8 m/s^2

6. Now, consider the forces acting on point B. There are two forces: the tension T2 acting vertically upward and the tension T1 acting horizontally to the left. Since point B is also in equilibrium, the vertical forces and horizontal forces must balance each other.

Vertical forces: T2 - T1 = 0

Horizontal forces: There are no horizontal forces acting on point B, so there is no horizontal component of tension T1.

7. Calculate the tension in the portion BC (T2). From the equation T2 - T1 = 0, we can solve for T2:

T2 = T1

T2 = (650g / 1000) kg × 9.8 m/s^2

Therefore, the tensions in the portions AC and BC of the string are equal and can be calculated using the formula (650g / 1000) kg × 9.8 m/s^2.