Question

You have developed a method in which a paint shaker is used to measure the coefficient of static friction between various objects and a known surface. the shaker oscillates with a fixed amplitude of 40 mm , but you can adjust the frequency of the motion. you have affixed a horizontal tabletop (the known surface) to the shaker so that the tabletop oscillates with it. then you put an object on the tabletop and increase the frequency until the object begins to slip on the surface. part a if a frequency f = 1.75 hz is required before a penny positioned on the tabletop starts to slide, what is the coefficient of static friction between penny and tabletop?

Answer 1

So we want find the coefficient of friction, which comes from F = (mu)*N, where F is the force of static friction, mu is the coefficient of friction and N is the normal force of the object. We can talk about how the penny would move after this force is reached with F = ma, where a is acceleration and m is the mass of the penny. The maximum acceleration of an oscillatory motion is a = Aω², where A is the amplitude and ω is the angular frequency. To turn ω into the regular frequency we're given, we can use ω = 2πf. Using this, our max acceleration becomes a = A(2πf)², which we can put into F=ma, so it becomes F = m(4Aπ²f²), then we can set it equal to F = (mu)N, so 4Amπ²f² = (mu)N. N, the normal force, will equal the weight force of the penny (draw a free body diagram), W = mg, where m is the mass of the penny and g is the gravitational constant (9.8m/s²). So N = W = mg, so we sub mg in for N, and we get 4Amπ²f² = (mu)mg, we solve for mu, cancelling our m's (good thing too since we don't know the mass of the penny) and we get mu = 4Aπ²f²/g, then we plug stuff in (convert mm to m by multiplying by 10⁻³) mu = 4(40*10⁻³)(3.14)²(1.75)²/(9.8) = 0.49 for our coefficient of friction. Lots of equations, but I hope that made sense!