Chapter 1 - Problems: Developing Engineering Skills - Page 30: 1.10

Here in this problem , use F = ma then , we take modulus on both sides, |F|= m|a| here, |a| = 60 g putting value of |a|, and mass as 50 lb, and using conversions, we get |F| = ( 60 )( 50 lb)( 32.2 ft/s) ( 1 lbf/( 32.2 lb.ft/$sec^{2}$ )=3000 lbf.

Here in this problem , use F = ma then , we take modulus on both sides, |F|= m|a| here, |a| = 60 g putting value of |a|, and mass as 50 lb, and using conversions, we get |F| = ( 60 )( 50 lb)( 32.2 ft/s) ( 1 lbf/( 32.2 lb.ft/$sec^{2}$ )=3000 lbf.