Given:
μ = 0.433
α = 30°
m
Find: Pa, Pb
a)
fkx+mgx+Nx+Px=0P=μmg=0.433 ∗ 10m/s2 ∗ m kg= 4.33m Nf_{k_x} + mg_x + N_x + P_x = 0\\ P = \mu mg = 0.433\ *\ 10m/s^2 \ *\ m\ kg =\ 4.33m\ Nfkx+mgx+Nx+Px=0P=μmg=0.433 ∗ 10m/s2 ∗ m kg= 4.33m N
b)
fkx+mgx+Nx+Px=0μmg=PcosαP=μmgcosα=0.433∗10m/s2 ∗ m kgcos30°≈5m Nf_{k_x} + mg_x + N_x + P_x = 0\\ \mu mg = Pcos\alpha\\ P = \dfrac{\mu mg}{cos\alpha} = \dfrac{0.433 * 10m/s^2\ *\ m\ kg}{cos 30\degree} \approx 5m\ Nfkx+mgx+Nx+Px=0μmg=PcosαP=cosαμmg=cos30°0.433∗10m/s2 ∗ m kg≈5m N
Answer: a) Pa=4.33m N b) Pb=5m N
(where m is the mass of a truck in kilos)
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