Problem 45 - An Introduction to the Mechanics of Solids in SI Units

Question

It is proposed to measure Newton's gravitational constant by hanging a 10 -m length of Mylar tape from a rod in a high ceiling and fastening the end to a 2 -m-length aluminum rod. On each end of the rod a cast-iron cannon ball of mass \(5 \mathrm{~kg}\) is fastened. Take the Mylar to have a tension modulus of \(1.4 \times\) \(10^{9} \mathrm{~N} / \mathrm{m}^{2}\), a Poisson's ratio of \(0.40\), and cross-sectional dimensions \(25.4\) by \(0.254 \mathrm{~mm}\). Verify that hanging the weights on the tape will not overload the Mylar, which has a maximum elastic strain capacity of 10 percent. When the whole device has stopped shaking and oscillating, the position of the balls is carefully determined, and then an aluminum bucket filled with lead shot is placed near each of the balls as shown. The distance between the center of the iron balls and the center of the buckets is \(400 \mathrm{~mm}\), and each bucket is filled with \(100 \mathrm{~kg}\) of lead shot. Estimate the expected deflection of each cannon ball if the gravitational attraction between two masses each of \(1 \mathrm{~kg}\) separated by \(1 \mathrm{~m}\) is \(6.67 \times 10^{-11} \mathrm{~N}\).

Step-by-Step Solution

Answer

The Mylar tape will not get overloaded as the resulting strain, 0.0054, is less than the maximum elastic strain capacity of 0.10 or 10 percent. Additionally, each cannon ball will deflect by about 0.9 mm.

Step 1: Calculate the Strain on Mylar Tape

To start with, calculate the force exerted by the iron balls on the Mylar tape: \( F = m \cdot g = 5 \cdot 9.8 = 49 \, N \), where \( m = 5 \, kg \) is the mass of the ball and \( g = 9.8 \, m/s^2 \) is the gravitational constant. Next, determine the cross-sectional area of the Mylar tape: \( A = (25.4 \times 10^{-3} m) \times (0.254 \times 10^{-3} m) = 6.45 \times 10^{-6} m^2 \). The stress in the tape is given by: \( \sigma = \frac{F}{A} = \frac{49}{6.45 \times 10^{-6}} = 7.6 \times 10^6 \, N/m^2 \). Finally, to verify that the strain in the Mylar tape is within the allowable limit, calculate the strain by dividing the stress by the tension modulus of the Mylar tape: \( \epsilon = \frac{\sigma}{E} = \frac{7.6 \times 10^6}{1.4 \times 10^9} = 0.0054 \). As this value is less than the maximum permissible strain of 0.10 (or 10 percent), the Mylar tape will not be overloaded.

Step 2: Estimate the Deflection of Each Cannon Ball

The force due to gravitational attraction between two masses is \( F = G \cdot \frac{m_1 \cdot m_2}{r^2} \), where \( G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2 \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of two bodies, and \( r \) is the distance between the centers of the two bodies. In this case, \( m_1 = 5 \, kg \) (mass of an iron ball this represents one of the bodies) and \( m_2 = 100 \, kg \) (mass of the lead shot in the bucket, which forms another body), and \( r = 400 \times 10^{-3} \, m \). Substituting these values, calculate the force: \( F = 6.67 \times 10^{-11} \cdot \frac{5 \cdot 100}{(400 \times 10^{-3})^2} = 8.34 \times 10^{-7} \, N \). This gravitational force induces tensile stress in the Mylar tape, which leads to a deflective movement of the cannon balls. Using the formula for stress (\( \sigma = \epsilon \times E \)), where \( E = 1.4 \times 10^9 \, N/m^2 \) is the tension modulus of the Mylar tape, the strain \( \epsilon = \frac{\sigma}{E} = \frac{F/A}{E} = \frac{8.34 \times 10^{-7}}{6.45 \times 10^{-6} \times 1.4 \times 10^{9}} = 9.09 \times 10^{-5} \). Then, use the strain to estimate the deflection in the Mylar tape: \( \Delta = l \cdot \epsilon = 10 \cdot 9.09 \times 10^{-5} = 0.000909 \, m \) or approximately 0.9 mm. Each cannon ball should be deflected by around 0.9 mm due to the gravitational pull of the lead-filled buckets.

Key Concepts

Newton's gravitational constant

Newton's gravitational constant, denoted by the symbol \(G\), is a fundamental constant in physics that appears in Newton's law of universal gravitation. It is a measure of the strength of gravity and plays a vital role in calculating the gravitational force between two masses. The value of \(G\) is approximately \(6.67 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2}\). This means that two 1 kg masses separated by 1 meter will experience a gravitational force of \(6.67 \times 10^{-11} \) N.

Understanding the gravitational constant is crucial in exercises dealing with gravity-related forces, as it allows us to compute the actual force experienced between objects of any mass separated by distance. For example, when estimating the deflection of cannon balls in an experiment using lead weights, \(G\) helps in calculating the tiny force due to gravitational attraction.

In practical terms, \(G\) is incredibly small, emphasizing the fact that gravitational forces are only significant for very massive bodies like planets and stars. Nevertheless, precise measurement of \(G\) remains of great scientific importance, as it helps in refining our understanding of fundamental physical laws.

Mylar tape mechanics

Mylar tape mechanics involve understanding how Mylar tape behaves under stress and its suitability for experimental or practical applications. Mylar is a brand of strong polyester film used in a variety of mechanical and electrical applications, known for its high tensile strength and thermal stability.

The exercise investigates the ability of a Mylar tape to support the weight of two 5 kg iron balls. The tape is expected to carry this load without exceeding its maximum elastic strain capacity, which is 0.10 or 10%. The tension modulus of the tape is given as \(1.4 \times 10^9 \, \mathrm{N/m^2}\), which indicates how much stress the material can withstand before deforming. Moreover, it has a Poisson's ratio of \(0.40\), reflecting how the tape will widen or narrow under tension.

The cross-sectional dimensions of the tape are \(25.4\) by \(0.254\) mm, equating to a cross-sectional area of \(6.45 \times 10^{-6} \, \mathrm{m^2}\). Calculating stress exerted on this area helps ensure that the tape is not overloaded. In this scenario, the computed stress of \(7.6 \times 10^6 \, \mathrm{N/m^2}\) results in a strain of \(0.0054\), significantly below the maximum threshold, proving the Mylar can safely bear the load.

Stress and strain analysis

Stress and strain analysis is pivotal in understanding how materials react when forces are applied. Stress is defined as force per unit area, and strain is the deformation experienced by the material due to this stress. In the given exercise, these concepts are critical in evaluating whether the Mylar tape can hold the weight of the iron cannon balls without yielding.

To begin with, the stress \( \sigma \) in the tape is calculated using the formula \( \sigma = \frac{F}{A} \), where \(F\) is the force due to gravity on the balls (\(49 \) N) and \(A\) is the cross-sectional area (\(6.45 \times 10^{-6} \, \mathrm{m^2}\)). The calculated stress verifies the tape will not be overloaded.

The strain \( \epsilon \) is then determined using the relation \( \epsilon = \frac{\sigma}{E} \), where \(E\) is the tension modulus of the Mylar. The resulting strain tells us how much the Mylar will stretch under the given stress, and in this case, it is within the permissible elastic limit.

Further analysis is used to estimate the deflection caused by gravitational forces from the lead buckets, calculated using the same principles. A small tensile stress leads to a very slight extension or movement of the material, showcasing the sophisticated interplay between stress and strain. These calculations ensure the structural integrity of materials in applications, even under small loads, allowing engineers to design safer and more efficient systems.