Understanding Frictional Force on an Inclined Plane

When a block rests on an inclined plane, several forces act on it. These forces include gravity, the normal force from the plane, and the frictional force.

Forces Acting on the Block

Let's analyze the forces acting on the block when it is on an inclined plane with an angle $\theta$ with the horizontal:

• Gravity ($mg$): Acts vertically downwards. This force can be resolved into two components:
  • Component parallel to the plane: $mg \sin(\theta)$ (pulling the block down the incline).
  • Component perpendicular to the plane: $mg \cos(\theta)$ (pressing the block into the incline).
• Normal Force ($N$): Acts perpendicular to the inclined plane, upwards. For a block resting on the plane, the normal force balances the perpendicular component of gravity: $N = mg \cos(\theta)$.
• Frictional Force ($f$): Acts parallel to the inclined plane, opposing the potential or actual motion. Its nature depends on whether the block is stationary or moving.

Static Friction while Stationary

The question states the block is "tying stationary". This means static friction is acting. While the block remains stationary on the inclined plane, the static frictional force must balance the component of gravity pulling the block down the slope.

Static friction $f_s$ is equal to $mg \sin(\theta)$.

So, $f_s = mg \sin(\theta)$.

As the angle $\theta$ of the inclined plane is slowly increased from zero, the value of $\sin(\theta)$ increases (for $0 \le \theta < 90^\circ$). Therefore, the static frictional force $f_s = mg \sin(\theta)$ must increase to keep the block stationary. This increase continues until the static friction reaches its maximum possible value for that angle.

Maximum Static Friction and Kinetic Friction

The maximum static frictional force depends on the static coefficient of friction ($\mu_s$) and the normal force ($N$).

$f_{s,max} = \mu_s N = \mu_s mg \cos(\theta)$.

As the angle $\theta$ increases, $\cos(\theta)$ decreases (for $0 \le \theta < 90^\circ$). This means the maximum possible static friction $f_{s,max}$ that the surface can provide actually decreases as the angle increases.

If the angle $\theta$ increases beyond the angle of repose (where $mg \sin(\theta) > f_{s,max}$), the block will start to slide. Once the block is sliding, the friction becomes kinetic friction.

The kinetic frictional force $f_k$ is given by:

$f_k = \mu_k N = \mu_k mg \cos(\theta)$

where $\mu_k$ is the kinetic coefficient of friction (usually $\mu_k < \mu_s$).

Notice that the kinetic frictional force also depends on $\cos(\theta)$. As the angle $\theta$ increases (while the block is sliding), $\cos(\theta)$ decreases, leading to a decrease in the kinetic frictional force.

Interpreting the Outcome

Based on the physics, while the block remains stationary, the actual static frictional force ($mg \sin(\theta)$) increases as $\theta$ increases. However, the maximum possible static friction ($\mu_s mg \cos(\theta)$) decreases as $\theta$ increases. Also, if the block starts sliding, the kinetic friction ($\mu_k mg \cos(\theta)$) decreases as $\theta$ increases.

Given the option suggesting the frictional force will "decrease", it implies either the question refers to the maximum possible static friction or the kinetic friction after the block has started sliding at a certain increased angle. While the initial static force increases, the capability of the surface to provide static friction decreases, and the friction during sliding also decreases with increasing angle. The provided answer aligns with the behavior of the maximum static friction limit or the kinetic friction once motion begins.