A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 1.86 Hz. On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is 5.20 10-2 m.
To obtain the coefficient of static friction between the tray and the cup, we can use the equation for simple harmonic motion, as well as the conditions for slipping.
1. Start by finding the angular frequency, ω, using the equation:
ω = 2πf
where f is the frequency given as 1.86 Hz.
Plugging in the values, we have:
ω = 2π(1.86 Hz)
2. Next, determine the amplitude of the motion, A, which is given as 5.20 * 10^-2 m.
3. The maximum acceleration, amax, is related to the angular frequency and amplitude of the motion.
amax = ω^2 * A
Substituting the values we found earlier, we have:
amax = (2π(1.86 Hz))^2 * (5.20 * 10^-2 m)
4. The maximum acceleration is also related to the static friction force, Fs, and the mass of the cup, m.
Fs = m * amax
Since the cup is empty, we assume its mass is negligible, so m = 0.
5. With m = 0, we can rewrite the equation as:
Fs = 0 * amax
Fs = 0
6. The static friction force, Fs, is zero when the cup begins slipping. Therefore, the coefficient of static friction, μs, is also zero.
μs = 0
Therefore, the coefficient of static friction between the tray and the cup is 0.
To obtain the coefficient of static friction between the tray and the cup, we can start by analyzing the forces acting on the cup.
In simple harmonic motion, the restoring force that brings the cup back to its equilibrium position is provided by the static friction between the tray and the cup. When the amplitude is reached, the cup begins to slip, indicating that the static friction has been overcome.
1. Determine the angular frequency (ω) of the simple harmonic motion:
The angular frequency (ω) can be calculated using the formula: ω = 2πf, where f is the frequency of the motion (given as 1.86 Hz).
Therefore: ω = 2π * 1.86 Hz = 11.67 radians/s.
2. Calculate the maximum acceleration (a_max) of the cup:
The maximum acceleration (a_max) is related to the angular frequency (ω) and amplitude (A) by the formula: a_max = ω^2 * A.
Given that the amplitude (A) is 5.20 * 10^-2 m, and ω = 11.67 radians/s, we can substitute these values into the formula to find a_max.
a_max = (11.67 rad/s)^2 * (5.20 * 10^-2 m) = 6.38 m/s^2.
3. Determine the force required to overcome the static friction:
The force required to overcome static friction is given by the formula: F_friction = m * a_max, where m is the mass of the cup.
Since the cup is empty, its mass is negligible, so we can assume m ≈ 0.
F_friction = 0 * 6.38 m/s^2 = 0 N.
4. Calculate the coefficient of static friction (μ_s):
The coefficient of static friction (μ_s) can be obtained by dividing the force of static friction (F_friction) by the force pressing the cup against the tray (which is equal to the weight of the cup in this case).
The weight of the cup (F_weight) can be calculated using the formula: F_weight = m * g, where g is the acceleration due to gravity.
Since the mass (m) is negligible, we can assume F_weight ≈ 0.
μ_s = F_friction / F_weight = 0 N / 0 N (which is an indeterminate form).
Since both the numerator and denominator in the equation are zero, the coefficient of static friction cannot be determined for this particular scenario.
x=Asin ωt
v=Aωcos ωt
a= Aω²sin ωt
F=ma=mAω²sin ωt.
F(max)=F(fr)
F(max)= mA ω²
F(fr) =μmg
mAω²=μmg
μ=A ω²/g=A(2πf)²/g =
=5.2•10⁻²(2π•1.86) ²/9.8=0.23