The brake shown in the following figure is applied by the spring and released by a hydraulic cylinder (Not Shown). Using "Short Shoe Drum Brake Theory" and a friction coefficient of μ=0.45 between the brake shoes and the drum, find the spring force "Fs" needed to produce a braking torque of Th = 1750 Nm and the magnitude of the force on the bearings supporting the brake drum at its center (Assume the weight of the brake drum is negligible.). The brake drum rotates in a clockwise direction. (Start by drawing "Free Body Diagrams" of the brake drum, the left arm/brake shoe assembly, and the right arm/brake shoe assembly. Be careful when assigning direction to the friction forces in the "Free Body Diagrams".) 500 mm nodon Spring 400 mm 300 mm de ono di S 350 mm Partial Ans. Fs 5824 Newtons

The figure shows an internal rim-type brake having an inside rim diameter of 300 mm and a dimension R 5 125 mm. The shoes have a face width of 40 mm and are both actuated by a force of 2.2 kN. The drum rotates clockwise. The mean coefficient of friction is 0.28. a. Find the maximum pressure and indicate the shoe on which it occurs. b. Estimate the braking torque effected by each shoe, and find the total braking torque. c. Estimate the resulting hinge-pin reactions. 30°- 30°. R 120° 120° Pin Pin 30° 30° Figure 3: An internal rim-type brake

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Form the figure shown, a flexible belt placed around a rotating drum of 4-inch radius acts as a brake when the arm ABCD is pulled down by the force P. The coefficient of kinetic friction between the belt and the drum is 0.7. What force P that would result in a braking torque of 400 lb . in., assuming that the drum is rotating Neglect the weight of the counterclockwise. brake arm. 60° 4 in. 60° A B C -2 in.²4in. 2 in. 42×-| -2 D P