91Ó°ÊÓ

Angle is given$\mathrm{θ}=30°$ .

The problem is based on Newton’s second law of motion which states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object. Draw free-body diagrams for each force separately and resolve the given force and weight. Using components of forces the net force equation can be written.

Formulae:

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{m}\mathbf{}\mathbf{a} \\ \mathbf{W}\mathbf{=}\mathbf{mg} \end{gathered}$$

The calculation for normal due to application of force a:

Due to the application of force, there is no motion in the vertical direction, so the net force is zero.

The equation for the net force can be written as,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=0 \\ {\mathrm{N}}_{\mathrm{a}}+\mathrm{a}\sin 30°-\mathrm{mgcos}30°=0 \\ {\mathrm{N}}_{\mathrm{a}}=\mathrm{mgcos}30°-\mathrm{asin}30° \\ {\mathrm{N}}_{\mathrm{a}}=0.866\mathrm{mg}-0.5\mathrm{a}\left(1\right) \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{normal}\mathrm{force}\mathrm{by}\mathrm{incline}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{application}\mathrm{of}\mathrm{force}\mathrm{a}. \end{gathered}$$

The calculation for normal due to application of force b:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=0 \\ {\mathrm{N}}_{\mathrm{b}}+\mathrm{b}\cos 30°-\mathrm{mgcos}30°=0 \\ {\mathrm{N}}_{\mathrm{b}}=\mathrm{mgcos}30°-\mathrm{asin}30° \\ {\mathrm{N}}_{\mathrm{b}}=0.866\mathrm{mg}-0.866\mathrm{b}\left(2\right) \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{normal}\mathrm{force}\mathrm{by}\mathrm{incline}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{application}\mathrm{of}\mathrm{force}\mathrm{b}. \end{gathered}$$

The calculation for normal due to application of force c:

Similarly, the net force equation for force c,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=0 \\ {\mathrm{N}}_{\mathrm{c}}+\mathrm{c}\cos 30°-\mathrm{mgcos}30°=0 \\ {\mathrm{N}}_{\mathrm{c}}=\mathrm{mgcos}30°-c\sin 30° \\ {\mathrm{N}}_{\mathrm{c}}=0.866\mathrm{mg}-0.5\mathrm{c}\left(3\right) \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{normal}\mathrm{force}\mathrm{by}\mathrm{incline}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{application}\mathrm{of}\mathrm{force}\mathrm{c}. \end{gathered}$$

The calculation for normal due to application of force d:

Similarly, the net force equation for force d,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=0 \\ {\mathrm{N}}_{\mathrm{d}}+\mathrm{d}\cos 30°-\mathrm{mgcos}30°=0 \\ {\mathrm{N}}_{\mathrm{d}}=\mathrm{mgcos}30°-d\sin 30° \\ {\mathrm{N}}_{\mathrm{d}}=0.866\mathrm{mg}+0.866\mathrm{d}\left(4\right) \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{normal}\mathrm{force}\mathrm{by}\mathrm{incline}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{application}\mathrm{of}\mathrm{force}\mathrm{d}. \end{gathered}$$

From equations (1), (2), (3), and (4) we get,

Forces a, b, c, and d have the same magnitude then the increasing order of normal forces is${\mathrm{N}}_{\mathrm{d}}>{\mathrm{N}}_{\mathrm{c}}>{\mathrm{N}}_{\mathrm{b}}>{\mathrm{N}}_{\mathrm{a}}$.