Answer
a) 1.6 $\times$ $10^{4}$ N
b) 4.3 $\times$ $10^{3}$ N
Work Step by Step
a) Normal force
= mg $\times$ cos(15)
= 1700 $\times$ 9.8 $\times$ cos(15)
$\approx$ 1.6 $\times$ $10^{4}$ N
b) Static frictional force
= mg $\times$ sin(15)
= 1700 $\times$ 9.8 $\times$ sin(15)
$\approx$ 4.3 $\times$ $10^{3}$ N
