How far below the drum center would the arm pivot need to


Figure P18.22 shows a brake with only one shoe, being applied by a 1.5-kN force. (The complete brake would normally have a second shoe in order to balance the forces, but only one shoe is considered here to keep the problem short.) Four seconds after force F is applied, the drum comes to a stop. During this time the drum makes 110 revolutions. Use the short-shoe approximation and an estimated coefficient of friction of 0.35.

(a) Draw the brake shoe and arm assembly as a free body in equilibrium.

(b) Is the brake self-energizing or deenergizing for the direction of drum rotation involved?

(c) What is the magnitude of the torque developed by the brake?

(d) How much work does the brake do in bringing the drum to a stop?

(e) What is the average braking power during the 4-second interval?

(f) How far below the drum center would the arm pivot need to be to make the brake self-locking for f = 0.35?

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Mechanical Engineering: How far below the drum center would the arm pivot need to
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