As the initially empty urinary bladder fills with urine and expands, its internal pressure increases by 3600 Pa, which triggers the micturition reflex (the feeling of the need to urinate). The horizontal square section of the bladder wall with an edge length of 0.012 m. Because the bladder is stretched, four tension forces of equal magnitude T act on the square section, one at each edge, and each force is directed at an angle è below the horizontal. What is the magnitude T of the tension force acting on one edge of the section when the uinternal bladder pressure is 3600 Pa and each of the four tension forces is directed 3.2° below the horizontal?

Question

As the initially empty urinary bladder fills with urine and expands, its internal pressure increases by 3600 Pa, which triggers the micturition reflex (the feeling of the need to urinate). The horizontal square section of the bladder wall with an edge length of 0.012 m. Because the bladder is stretched, four tension forces of equal magnitude T act on the square section, one at each edge, and each force is directed at an angle è below the horizontal. What is the magnitude T of the tension force acting on one edge of the section when the uinternal bladder pressure is 3600 Pa and each of the four tension forces is directed 3.2° below the horizontal?

Answer 1

To find the magnitude T of the tension force acting on one edge of the section, we can use the concept of equilibrium. In equilibrium, the sum of the forces in each direction (horizontal and vertical) is zero.

Let's break down the forces acting on the square section:

1. Tension force directed horizontally towards the right (T*cos(θ))
2. Tension force directed horizontally towards the left (T*cos(θ))
3. Tension force directed vertically downwards (T*sin(θ))
4. Tension force directed vertically upwards (T*sin(θ))
5. Pressure force acting perpendicular to the square section (P * A)

Since the square section is in equilibrium, the sum of the horizontal forces is zero:

T*cos(θ) - T*cos(θ) = 0

This implies that T*cos(θ) = T*cos(θ).

Similarly, the sum of the vertical forces is zero:

T*sin(θ) + T*sin(θ) - P*A = 0

Simplifying this equation, we get:

2*T*sin(θ) = P*A

Now we can substitute the given values:

θ = 3.2° (converted to radians)
P = 3600 Pa
A = (0.012 m)^2

Using these values, we can solve for T:

2*T*sin(3.2°) = 3600 Pa * (0.012 m)^2

Simplifying further, we have:

T*sin(3.2°) = 3600 Pa * (0.012 m)^2 / 2

Now, divide both sides by sin(3.2°):

T = (3600 Pa * (0.012 m)^2 / 2) / sin(3.2°)

Calculating this expression will give us the magnitude T of the tension force acting on one edge of the section when the internal bladder pressure is 3600 Pa and each of the four tension forces is directed 3.2° below the horizontal.