Answer:
The magnitude of the minimum normal force that the brake shoe exerts on the shaft is approximately 5.623 Newtons.
Step-by-step explanation:
To find the magnitude of the minimum normal force that the brake shoe exerts on the shaft, we need to consider the forces acting on the rotating coil and use the conditions for static equilibrium.
The forces acting on the rotating coil are as follows:
Magnetic torque due to the interaction between the magnetic field and the current-carrying coil, which tries to rotate the coil.
Frictional force exerted by the brake shoe on the shaft, which opposes the rotation of the coil and keeps it from turning.
For static equilibrium, the net torque acting on the coil must be zero, and there should be no angular acceleration. The magnetic torque is given by:
Magnetic Torque (τ) = N * B * A * I * sin(θ)
where:
N = Number of turns in the coil = 520
B = Magnetic field strength = 0.10 T
A = Area per turn of the coil = 2.4 x 10^(-3) m^2
I = Current in the coil = 0.13 A
θ = Angle between the magnetic field and the normal to the coil's plane = 90° (perpendicular)
τ = 520 * 0.10 T * 2.4 x 10^(-3) m^2 * 0.13 A * sin(90°)
τ = 520 * 0.10 * 2.4 x 10^(-3) * 0.13
Next, we need to calculate the frictional torque exerted by the brake shoe on the shaft. The coefficient of static friction (μ) between the shaft and the brake shoe is given as 0.74. The frictional torque (τ_f) is given by:
Frictional Torque (τ_f) = μ * R * F_N
where:
μ = Coefficient of static friction = 0.74
R = Radius of the shaft = 0.0078 m
F_N = Normal force exerted by the brake shoe on the shaft (what we want to find)
Now, since the coil is in static equilibrium, the magnetic torque and the frictional torque must be equal:
τ = τ_f
520 * 0.10 * 2.4 x 10^(-3) * 0.13 = 0.74 * 0.0078 * F_N
Now we can solve for the normal force (F_N):
F_N = (520 * 0.10 * 2.4 x 10^(-3) * 0.13) / (0.74 * 0.0078)
F_N ≈ 5.623 N
So, the magnitude of the minimum normal force that the brake shoe exerts on the shaft is approximately 5.623 Newtons.