Numericals: System of Forces | Engineering Mechanics - Civil Engineering (CE) PDF Download

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:From relationship 1/3, we have

W = mg = (95.9)(32.2) = 3090 lb Ans.

From the table of conversion factors inside the front cover of the textbook, we see that 1 slug is equal to 14.594 kg. Thus, the mass of the car in slugs is

m = 1400 kg*( 1 slug/ 14.594 kg) = 95.9 slugs Ans.

Finally, its weight in pounds is

W = mg = 1400(9.81) = 13 730 N Ans.

As another route to the last result, we can convert from kg to lbm. Again using the table inside the front cover, we have

The weight in pounds associated with the mass of 3090 lbm is 3090 lb, as calculated above. We recall that 1 lbm is the amount of mass which under standard conditions has a weight of 1 lb of force. We rarely refer to the U.S. mass unit lbm in this textbook series, but rather use the slug for mass. The sole use of slug, rather than the unnecessary use of two units for mass, will prove to be powerful and simple—especially in dynamics.

Problem 2: Use Newton’s law of universal gravitation to calculate the weight of a 70-kg person standing on the surface of the earth. Then repeat the calculation by using Wmg and compare your two results. Use Table D/2 as needed.

Solution 2: The two results are

W = G(me)m/ R^2 = (6.673*10^(-11))(5.976*10^24)(70)/ [6371*1000]^2 = 688 N   Ans.

W = mg = 70(9.81) = 687 NAns.

The discrepancy is due to the fact that Newton’s universal gravitational law does not take into account the rotation of the earth. On the other hand, the value g 9.81 m/s2 used in the second equation does account for the earth’s rotation. Note that had we used the more accurate value g = 9.80665 m/s2 (which likewise accounts for the earth’s rotation) in the second equation, the discrepancy would have been larger (686 N would have been the result).