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Torque and hydraulic brake system question. (Homework help.) :)?

Figure P9.25 shows the essential parts of a hydraulic brake system. The area of the piston in the master cylinder is 1.75 cm2, and that of the piston in the brake cylinder is 6.4 cm2. The coefficient of friction between shoe and wheel drum is 0.50. If the wheel has a radius of 32 cm, determine the frictional torque about the axle when a force of 39 N is exerted on the brake pedal.To see image I will email it to you.I looked on yahoo answers before and someone gave answers that didn‘t make sense to me. They said to find the normal force (Fn) to find the force of friction (Ff) and that the torque (T) will be equal to Ff*r*(sin theta).They said Fn Force exerted on brake and that Ff Fn* coefficient of friction/(area of master cylinder + area of brake cylinder).It seems to me that they are confusing pressure with force. Because doesn‘t PressureForce/Area??? Please explain how to use the radius of each cylinder and why we use it in that way.

Answer:

As Fred says, Citroen hydraulic systems use green LHM fluid, as do any others with Citroen-based hydraulic suspension systems jointly with Rolls Royce. that is unique and would not mixture with the different fluid. LHM absorbs in basic terms an infinitesimal share of moisture, in comparison to widely used brake fluid. (in case you look on the suspension spheres in a Rolls Royce boot you will see the Citroen chevrons!)
Hydraulic Brake Cylinder
Well, F1/ A1(Master Cylinder) F2/A2(brake cylinder) base on Pascal Principle the force is exacter on the brake pedal, which mean it is F1(on master Cylinder becasue they are connected) They give you Area for each cylinder and force exert on brake(force on master cylinder),so:. F2 F1 * A2/ A1 39* 6.4/ 1.75 142.63 N Normal Force force exert on brake(that's true, refer back to the picture and that force is perpendicular to the surface of wheel drum) So frictional torque Ff * x Fn * Coefficient of friction * 0.32(radius) 142.63 * 0.5 * 0.32 22.82 N*m

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