Animated Solution for Physics - Laws of Motion: What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move? (2003)

Why Angle Matters in Friction

Imagine you are trying to move a heavy box across a rough floor. Intuition tells you to push it forward. But what happens if you push it downwards at an angle? This classic physics problem explores the hidden consequences of angled forces and how they interact with friction.

When a force $F$ is applied at an angle of $60^{\circ }$ downwards relative to the horizontal, it doesn't just act in one direction. To understand its true effect, we must break it down into its rectangular components.

The horizontal component, $F\cos 60^{\circ }$, is the driving force. This is the part of your push that is actively trying to slide the block across the floor. However, the vertical component, $F\sin 60^{\circ }$, acts straight down. This downward push is the trap! It effectively adds to the block's weight, pressing it harder against the surface.

Because the block is not sinking into the floor, it must be in vertical equilibrium. The upward normal force ($N$) exerted by the floor must perfectly balance all the downward forces. Therefore, the normal force is not just the weight of the block ($mg$); it is the weight plus the downward component of your push:

This increased normal force is crucial because the maximum static friction available to resist motion is directly proportional to it ($f_{\text{max}}=\mu N$). By pushing downwards, you are inadvertently increasing the friction you have to overcome!

For the block to remain at rest, the horizontal driving force must be less than or equal to the maximum static friction. We can write this condition as:

Substituting our expression for the normal force, we get the master inequality that governs the entire system:

Now, let's plug in the specific values given in the problem: $m=\sqrt{3}$ kg, $\mu =\frac{1}{2\sqrt{3}}$, $\cos 60^{\circ }=\frac{1}{2}$, and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$.

At first glance, this looks like a messy equation filled with square roots. But watch what happens when we distribute the coefficient of friction. The $\sqrt{3}$ in the denominator perfectly cancels out the $\sqrt{3}$ in both terms inside the parentheses!

The Final Threshold

With the roots gone, the algebra becomes incredibly simple. We group the $F$ terms by subtracting $\frac{F}{4}$ from both sides:

Multiplying both sides by 4 gives us the final, elegant result:

Given that the acceleration due to gravity $g=10{\text{ m/s}}^2$, the maximum force $F$ that can be applied without moving the block is $20\text{ N}$. Any force greater than this will overcome the static friction, and the block will begin to slide.