Step-by-step explanation:
To find the force required to pull the triangular shaft, we need to use the formula for the frictional force in a lubricated bearing:
F = μ * A * P / d
where F is the frictional force, μ is the viscosity of the lubricating oil, A is the area of contact between the shaft and the bearing, P is the pressure exerted by the shaft on the bearing, and d is the thickness of the oil film.
In this case, we can assume that the pressure is uniform across the contact area and that the oil film thickness is equal to the average of t1, t2, and t3, which is (1+1+1)/3 = 1 mm = 0.001 m.
The area of contact can be calculated as the perimeter of the shaft multiplied by the length of the bearing:
A = l * (t1 + t2 + t3) = 0.1 m * (0.001 m + 0.001 m + 0.001 m) = 0.0003 m^2
To find the pressure, we need to consider the weight of the shaft and any external forces acting on it. Since the shaft is pulled at a constant velocity, the external force required to overcome the frictional force must be equal and opposite to the frictional force. Therefore, we can set the frictional force equal to the weight of the shaft:
F = m * g
where m is the mass of the shaft and g is the acceleration due to gravity.
The mass of the shaft can be calculated as the product of its density and volume:
m = ρ * V
where ρ is the density of the shaft material (assumed to be uniform) and V is its volume. Since the shaft is triangular in shape, we can use the formula for the volume of a triangular prism:
V = l * (t1 + t2 + t3) * h / 2
where h is the height of the triangle, which we can assume to be equal to the average of t1, t2, and t3 (since the triangle is equilateral):
h = (t1 + t2 + t3) / 3 = 0.001 m
Substituting the given values, we get:
V = 0.1 m * (0.001 m + 0.001 m + 0.001 m) * 0.001 m / 2 = 1.5 x 10^-7 m^3
Assuming that the shaft material is steel, with a density of 7850 kg/m^3, we get:
m = 7850 kg/m^3 * 1.5 x 10^-7 m^3 = 1.1775 x 10^-3 kg
Substituting the values for A, d, m, and g into the formula for the frictional force, we get:
F = μ * A * P / d = m * g
μ * A * P / d = m * g
P = m * g * d / (μ * A)
P = 1.1775 x 10^-3 kg * 9.81 m/s^2 * 0.001 m / (1 x 10^-1 Ns/m^2 * 0.0003 m^2) = 130.833 N/m^2
Finally, we can calculate the frictional force by multiplying the pressure by the area of contact:
F = P * A = 130.833 N/m^2 * 0.0003 m^2 = 0.03925 N
Therefore, the force required to pull the triangular shaft at a