What Determines the Coefficient of Static Friction in Oscillating Blocks?

[Havok104]

Homework Statement

The two blocks in the figure oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the amplitude is increased to 36 cm.

What is the coefficient of static friction between the two blocks?

Homework Equations

The Attempt at a Solution

I've tried to think of the forces involved (ie. friction -normal force * coefficient of friction etc)
But I'm having trouble incorporating the period and the amplitude into equations with forces in them.

Any help is greatly appreciated!

 

[tiny-tim]

Welcome to PF!

Hi Havok104! Welcome to PF!

Hint: call the coefficient of static friction µ.

What is the maximum acceleration that the upper block can have without slipping?

 

[kerimek]

I would approach this problem by first identifying the key variables and parameters involved. The period of oscillation and the amplitude of oscillation are given, and the question asks for the coefficient of static friction between the two blocks.

To solve this problem, we can start by considering the forces acting on the upper block. The only horizontal force acting on the upper block is the static friction force from the lower block. This friction force must be equal to the centripetal force needed to keep the upper block in circular motion. Therefore, we can use the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the upper block, v is its velocity, and r is the radius of the circular motion (in this case, the amplitude of oscillation).

We also know that the period of oscillation is related to the velocity and amplitude by the equation T = 2πr/v. Rearranging this equation, we get v = 2πr/T.

Now we can substitute this expression for velocity into the centripetal force equation to get Fc = m(2πr/T)^2/r. Simplifying, we get Fc = 4π^2mr/T^2.

Next, we can consider the forces acting on the lower block. It is in contact with the surface, so there must be a normal force acting on it. This normal force must be equal in magnitude to the weight of the upper block, since the two blocks are in equilibrium. Therefore, we can write Fc = mg, where g is the acceleration due to gravity.

Finally, we can equate the two expressions for Fc and solve for the coefficient of static friction, μs. This gives us μs = 4π^2m/T^2g. Plugging in the given values of the period and the mass of the upper block, we can calculate the coefficient of static friction between the two blocks.