Reconsider Question 6 but with the force $\overrightarrow{\mathbf{f}}$now directed down the ramp. As the magnitude of $\overrightarrow{\mathbf{f}}$is increased from zero, what happens to the direction and magnitude of the frictional force on the block?

a) A block is stationary with the frictional force acting on it.

b) A force-directed down the ramp is applied and gradually increased from zero.

To find the effect on the frictional force, we have to draw the free body diagram and then use Newton's 2nd law of motion. Applying this concept to the free-body diagram, we can get the behavior of the force.

Formulae:

The force due to Newton’s second law,

F =ma (1)

The force due to gravitational weight,

F =mg (2)

The static frictional force acting on the body,

${\mathbf{f}}_{\mathbf{s}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{s}}{\mathbf{F}}_{\mathbf{N}}$ (3)

Free body diagram of blocks:

When F=0 N the block is stationary, then the net force is 0 N so to balance the downward force due to gravity from equation (2),Mgsin$\mathrm{\left(}\mathrm{\theta }\mathrm{\right)}$,$f_s$is along the vertical direction, i. e., along the inclination upward direction (As shown in the FBD1).

If we increase the force F from 0 N , then the box will move along the inclination then to oppose this motion${\mathrm{f}}_{\mathrm{s}}$is upward (As shown in the FBD2).

By using Newton’s 2nd law of motion along the y-direction, the net forces acting on the block using equation (1) and equation (2) as follows:

(Block is not in motion along the vertical direction so,$a_y=0$)

$$\begin{gathered} \sum _{}\mathrm{F}={\mathrm{ma}}_{\mathrm{y}} \\ \mathrm{N}-\mathrm{Mg}\cos \left(\mathrm{\theta }\right)=0 \\ \mathrm{N}-\mathrm{Mg}\cos \left(\mathrm{\theta }\right) \end{gathered}$$

From this, we can say that N it only depends on the mass, and the inclination angle$\theta$ which is constant in this case so N remains the same.

And, the frictional force on the body using equation (3) is given as:

$$\begin{gathered} {\mathrm{f}}_{\mathrm{s}}=\mathrm{\mu N} \\ =\mathrm{\mu Mgcos\theta } \end{gathered}$$

From this${\mathrm{f}}_{\mathrm{s}}$ , it only depends on the normal force, N. As the normal force N is not changing the magnitude of ${\mathrm{f}}_{\mathrm{s}}$it remains the same.