Minimum force to move a block

Problem

A block of mass $m = 10\ \text{kg}$ is at rest on the ground. The coefficient of static friction between block and ground is $\mu_s = 0{,}40$. What is the minimum force required to set it moving if applied horizontally? And if the same force is applied at $30^\circ$ above the horizontal?

Solution

The block starts moving when the applied force equals the maximum value of static friction, $F_{a,\max} = \mu_s N$.

Case 1: horizontal force.

In the vertical direction the block remains at rest, so the normal simply balances the weight: $N = mg$

The minimum force to set the block moving equals the maximum static friction: $F = \mu_s\, mg = 0{,}40 \cdot 10 \cdot 9{,}81 = 39{,}24\ \text{N}$

$\ev{F_{\text{horiz}} \approx 39{,}2\ \text{N}}$

Case 2: force inclined at $30^\circ$ above the horizontal.

Now the force has a vertical component $F\sin\theta$ directed upward, which lightens the block. The vertical equilibrium becomes: $N = mg - F\sin\theta$

The condition for slipping concerns the horizontal components: the component $F\cos\theta$ must equal the maximum friction $\mu_s N$: $F\cos\theta = \mu_s\,(mg - F\sin\theta)$

We collect the terms in $F$: $F\cos\theta + \mu_s F\sin\theta = \mu_s\, mg$

$F = \frac{\mu_s\, mg}{\cos\theta + \mu_s\sin\theta}$

We substitute $\cos 30^\circ = 0{,}866$ and $\sin 30^\circ = 0{,}500$: $F = \frac{39{,}24}{0{,}866 + 0{,}40 \cdot 0{,}500} = \frac{39{,}24}{1{,}066}$

$\ev{F_{30^\circ} \approx 36{,}8\ \text{N}}$

Tilting the force upward reduces the intensity required: the vertical component unloads the block and lowers the friction to be overcome.

Links

Topics: Friction · Dynamics Concepts: Static friction · Newton’s second law Skills: Vector resolution Objects: Block