Numericals System of Forces - Engineering Mechanics - Civil Engineering (CE)

Problem1: Determine the weight in newtons of a car whose mass is 1400 kg. Convert the mass of the car to slugs and then determine its weight in pounds.

Solution1:

The weight of a body in SI units is given by the product of its mass and the acceleration due to gravity.

W = m·g

W = 1400 · 9.81 N/kg·m/s²

W = 13734 N

Ans. 13 730 N (rounded to three significant figures)

To obtain the mass in the imperial mass unit slug, use the conversion factor 1 slug = 14.594 kg.

m (in slugs) = 1400 kg · (1 slug / 14.594 kg)

m (in slugs) = 95.9 slug

Ans. 95.9 slugs

To compute weight in pounds (lb) using the slug as the mass unit, multiply mass in slugs by the gravitational acceleration in ft/s² (commonly approximated as 32.2 ft/s² in many engineering texts).

W = m (slugs) · g (ft/s²)

W = 95.9 · 32.2

W = 3090 lb

Ans. 3090 lb

Alternative route (relating SI mass to U.S. customary mass unit lbm): using conversion tables, a mass in kilograms can be converted to pounds-mass (lbm), and under standard conditions 1 lbm has a weight of 1 lb (force). The numerical result for weight in pounds obtained through the slug route is consistent with the usual kg → lbm → lb conversion. Note that this textbook prefers the slug as the imperial mass unit because using slugs keeps the relationship W = m·a simple in imperial units.

Solution 2:

Using Newton's law of universal gravitation, the gravitational force (weight) on a mass m at the surface of the Earth is

W = G·M_e·m / R^2

Substitute the commonly used values:

G = 6.673 × 10^-11 N·m²/kg²

M_e = 5.976 × 10²⁴ kg

R = 6371 × 10³ m

m = 70 kg

W = (6.673 × 10^-11) · (5.976 × 10²⁴) · 70 / (6371 × 10³)²

W = 688 N

Ans. 688 N

Using W = m·g with the standard gravitational acceleration used in many engineering problems:

W = 70 · 9.81

W = 687 N

Ans. 687 N

The small discrepancy between the two results arises because the value of g = 9.81 m/s² commonly used in W = m·g already incorporates the effect of the Earth's rotation (the centrifugal effect reduces the effective weight slightly) and local variations in g. The Newtonian universal-gravitation calculation shown above gives the force due purely to gravitational attraction between the Earth and the person, not including the rotational (centrifugal) correction. If one uses the more precise conventional value g = 9.80665 m/s² (which likewise accounts for rotation and is an internationally standard reference value), the numeric W = m·g would change slightly (for example, to about 686 N) and the difference compared with the pure gravitational calculation would be slightly larger.

Remarks and practical notes:

When solving weight and mass conversions in engineering problems, keep the following points in mind:

1. Use SI units (kg, m, s) when computing weight in newtons; apply W = m·g with g ≈ 9.81 m/s² unless a more precise local value is required.
2. For U.S. customary calculations, use the slug as the mass unit so that W (in pounds) = mass (in slugs) × g (≈ 32.2 ft/s²).
3. The unit lbm (pound-mass) is sometimes used in U.S. practice; 1 lbm has a weight of 1 lbf under standard gravity. To avoid confusion in dynamics problems, the slug is often preferred because of its direct relation to lbf through acceleration units.
4. Newton's law of universal gravitation provides the theoretical weight due to attraction between masses; practical values of g used in W = m·g include local and rotational corrections and are therefore suitable for routine engineering calculations.