# Monthly Archives: July 2009

## Spring Pushing a Block Off Another Block Involving Simple Harmonic Motion

Two blocks, the small one of mass 3.5 kg and the large one of mass 7.3 kg are stacked, small block on top. The large block is connected to a horizontal spring of spring constant 210 N/m. The large block sits on a frictionless floor. The coefficient of static friction between the two blocks is 0.492. What amplitude of simple harmonic motion of the spring-blocks system puts the smaller block on the verge of slipping over the larger block?

The maximum acceleration is what is important here. The simple harmonic motion is merely a spin on an old problem. So lay out some Newtonian Force equations: $F=ma$ – Standard Force Equation $F_{s} = -kx$ – Hook’s Law for Spring Motion $F_{N} = mg$ – Normal Force $F_{f} = \mu_{s} F_{N}$ – Frictional Force, tailored for the small block.

Start with the Spring equation, treating it as an external force acting on the two block system: $F_{s} = -kx$

Because it is the singular external force acting on the system, we can set the general force equation equal to this, reminding ourselves that we must add the two block masses together as they are one system currently: $M$ – Mass of the small block and the mass of the large block. $-kx = Ma$ $\frac{-kx}{M} = a$ – Acceleration of the two block system.

With that, we concentrate on the two block system internally now. We find the maximum force that friction exerts between the two blocks: $m$ – Mass of the small block. $F_{f} = \mu_{s} F_{N}$ – Frictional Force $F_{f} = \mu_{s} mg$ – Substitute Normal Force $F_{f} = \mu_{s} mg = ma$ – Friction is the only force acting on the block. $F_{f} = \mu_{s} g = a$ – Small mass cancels

Substitute one acceleration for the other: $\mu_{s} g = \frac{-kx}{M}$

Solve for x: $x = \frac{M\mu_{s}g}{-k}$

We can solve directly for x here because we know that the maximum acceleration that the system can undergo without sliding is always the negative of the maximum x value. This is a property of simple harmonic motion. $|x| = x_{max}$

And we solve: $x = \frac{(3.5[kg] + 7.3[kg]) * 0.492 * 9.8[m/s^2]}{-210[N/m]}$ $x_{max} = 0.248[m]$

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