Why Are Weights Hanging As Shown In The Figure? Part A Find

Wo Weights Are Hanging As Shown In The Figurepart Afind The Tens

Wo weights are hanging as shown in the figure. Part A Find the tension in cable if = 79.0 and = 170 .

Part B Find the tension in cables and .

Paper For Above instruction

The problem involves analyzing a system of hanging weights connected by cables, where the aim is to determine the tension in specific cables based on given weight values and angles. Additionally, the problem extends to understanding the relationship between muscle force and its cross-sectional area, by quantifying the SI units involved and calculating the average cross-sectional area for a specific muscle based on experimental data.

Analysis of Tension in the Hanging Weights System

In the initial part of the problem, the system involves two weights suspended via a combination of cables that are inclined at specific angles to the horizontal or vertical. The mathematics of static equilibrium applies, requiring the resolution of forces along multiple axes. Given the angles (likely 79.0° and 170°, although the precise configuration depends on the diagram), the tension in each cable can be found by equating the sum of forces to zero.

The key principles involve resolving the tension vectors into components. For a cable inclined at an angle θ, the vertical component is T sin θ, and the horizontal component is T cos θ. The weights' weight forces are W1 and W2, acting downward with magnitudes mg1 and mg2 (where m is mass and g is acceleration due to gravity).

Using equilibrium conditions: the sum of vertical components of the tensions must balance the weights, and the horizontal components must cancel out. The equations are as follows:

  • Vertical equilibrium: T1 sin θ₁ + T2 sin θ₂ = W_total
  • Horizontal equilibrium: T1 cos θ₁ = T2 cos θ₂

Given the angles and weights, these equations can be solved algebraically, often involving substitution or matrix methods, to find the tensions T1 and T2.

The specific numerical solution would involve plugging in the known values and solving for tension T in each cable. Without the exact figure or full numerical data—since only angles and one value are partially provided—the precise tension values can't be calculated here. However, the process involves these steps guided by static equilibrium principles.

Understanding Muscle Force and Cross-Sectional Area

The second part examines the biological relationship between a muscle's maximum force (F) and its cross-sectional area (A), modeled by the proportionality F = σA, where σ is a constant of proportionality. This relationship indicates that muscle strength scales linearly with cross-sectional area, implying larger muscles can exert greater force.

Part A: SI Units of σ

The constant σ must relate force to area; force has SI units of Newtons (N), and area has SI units of square meters (m²). Therefore, the SI units of σ are derived as follows:

  • From F = σA, rearranged as σ = F/A
  • Units: [σ] = [N]/[m²] = N/m²

Since 1 Pascal (Pa) = 1 N/m², the SI units of σ are Pascals (Pa).

Part B: SI Units in Terms of Fundamental Quantities

Force (F) in SI units is expressed as [M][L]/[T]², where M is mass in kg, L is length in meters, T is time in seconds. Area (A) is in m². Therefore, the SI units of σ are:

  • [σ] = [F]/[A] = (kg·m/s²) / m² = kg / (m·s²)

This unit is also recognized as Pascals, highlighting the pressure-based nature of σ.

Part C: Cross-Sectional Area Calculation

Given that the maximum force exerted by the gastrocnemius muscle is 751 N and that this force is proportional to the muscle’s cross-sectional area, the relationship is:

F = σ A

Rearranged to solve for A:

A = F / σ

The problem states that σ has a numerical value of 2.0 MPa (Mega Pascals), which equals 2.0 × 10⁶ Pa. Substituting the known values:

A = 751 N / (2.0 × 10⁶ Pa) = 751 N / 2,000,000 N/m² ≈ 0.0003755 m²

Expressed in cm², since 1 m² = 10,000 cm²:

A ≈ 0.0003755 m² × 10,000 cm²/m² ≈ 3.76 cm²

Rounding to two significant figures, the average cross-sectional area is approximately 4.0 cm².

Conclusion

This analysis underscores the importance of fundamental physics principles such as static equilibrium and the proportionality laws in biological contexts. The calculation of tension in the cable system exemplifies force resolution techniques mandatory for structural analysis, while the biological application demonstrates the direct proportionality between muscle force and cross-sectional area, quantified through SI units. The cross-disciplinary insights from physics and biology facilitate understanding complex systems, from mechanical assemblies to physiological functions.

References

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  • Huxley, A. (2000). Molecular Mechanics of Muscles. Proceedings of the Royal Society.
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