Given,
The angle in which the force is applied, θ=30°
The weight of the block, W=20 N
The applied force, F_app=10.0 N
As the block is in vertical equilibrium, the net vertical force on the block is zero.
Therefore,
[tex]\begin{gathered} N+F\sin \theta=W \\ \Rightarrow N=W-F\sin \theta \end{gathered}[/tex]
Where N is the normal force acting on the block.
On substituting the known values,
[tex]\begin{gathered} N=20-10\sin 30\degree \\ =15\text{ N} \end{gathered}[/tex]
Thus the magnitude of the normal force acting on the block is 15 N
