1. 49. We denote the mass of the slab as m, its density as ρ, and volume as V . The angle of inclination is
θ = 26◦ .
(a) The component of the weight of the slab along the incline is
F1 = mg sin θ = ρV g sin θ
(3.2 × 103 kg/m )(43 m)(2.5 m)(12 m)(9.8 m/s2 ) sin 26◦ = 1.77 × 107 N .
3
=
(b) The static force of friction is
fs = µs N = µs mg cos θ = µs ρV g cos θ
(0.39)(3.2 × 103 kg/m )(43 m)(2.5 m)(12 m)(9.8 m/s2 ) cos 26◦ = 1.42 × 107 N .
3
=
(c) The minimum force needed from the bolts to stabilize the slab is
F2 = F1 − fs = 1.77 × 107 N − 1.42 × 107 N = 3.5 × 106 N .
2
If the minimum number of bolts needed is n, then F2 /nA ≤ 3.6 × 108 N/m , or
3.5 × 106 N
n≥ 2 = 15.2 .
(3.6 × 108 N/m )(6.4 × 10−4 m2 )
Thus 16 bolts are needed.