A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the smallest value of mass 'm' for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope. в 27° A D 50 kg m E
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- Find the largest value of b/h at which the folding table is in equilibrium. The coefficients of static friction are 0.5 at A and 0.3 at C. Neglect the weight of the table.A clockwise couple M is applied to the circular cylinder as shown. Determine the value of M required to initiate motion for the conditions mg = 4.6 kg, mc - 4.7 kg, (s)B=0.46, (c=0.34, and r=0.21 m. Friction between the cylinder C and the block B is negligible. (s)c Answer: M = mc M mg (₂)B N.mDetermine the minimum and maximum values of the force (P) to keep the equilibrium for the block with a mass of weight 350 N resting on a rough surface inclined at an angle of 25°with the horizontal. Knowing that the angle of friction is 40°.
- 20 cm The homogeneous block weights 250 N. Determine the value of P for which the block 30 cm in equilibrium, the coefficient of friction is 0.3 between the plane and the block. 15 cm 10 N 10 cmAs shown, a man is leaning against the side of a cabinet with an unusual design. The cabinet's main body weighs 25 kg, while the upper rectangular portion weighs 3 kg. Assume the coefficients of friction between the cabinet and the floor are μs = 0.33 and μk = 0.28. Knowing that the force P exerted by the man's shoulder on the horizontal cabinet: 2. Determine which of the following is the CORRECT equilibrium equation obtained from the system's force diagram. A. ΣF = 0: N - 245.25 = 0B. ΣF = 0: P - μ N = 0C. ΣM = 0: 245.25(0.55) + 29.43(0.2) - N(x) - P(1.5) = 0A pair of wedges is used to lift a crate as shown in The crate weighs 3000 N, the wedge angle 0 is 15°, and the coefficient of friction is 0.25 at all surfaces. The weight of the wedges is negligible. Determine Fig. If the system would were removed. be in equilibrium if the force P W
- A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. If h = 32 in., determine the magnitude of the force P required to move the cabinet to the right (a) if all casters are locked, (b ) if the casters at B are locked and the casters at A are free to rotate, (c) if the casters at A are locked and the casters at B are free to rotate.1.2 kN H, = 0.35 PROBLEM 8.1 HE = 0.25 Determine whether the block shown is in equilibrium and find the magnitude and P direction of the friction force when 0= 25° and P = 750 N.A pair of wedges is used to lift a crate as shown in The crate weighs 4000 lb, the wedge angle e is Fig. 18°, and the coefficient of friction is 0.15 at all surfaces. The weight of the wedges is negligible. Determine The maxımum angle 8 for which the system would be in equilibrium if the force P were removed. W
- Q.4) Determine the maximum mass m of cylinder for which the system shown below is in Static equilibrium. The static coefficient of friction between the 50-kg block and the inclined plane is 0.15, and that between the cord and cylindrical disk support is 0.25. PRIN μ = 0.15 50 kg 20° μ = 0.25 mQ4.) An 8-foot-long uniform slender rod is connected to a 20-lb. block by a cord and pulley system as shown where LAO = 6 ft. The coefficient of static friction between the block and the surface is 0.6. Determine the range of rod weights for which the system remains in equilibrium. Assume friction is negligible between the pulleys and the cord. Consider both slipping and tipping. 30° 30° 6 ft 2.5 ft/ B 2 ft 40⁰ 3.5 ftThe axle of the pulley is frozen and cannot rotate with respect to the block. Knowing that the coefficient of static friction between cable ABCD and the pulley varies between 0.0 and 0.55, determine (a) the corresponding values of 0 for the system to remain in equilibrium, (b) the corresponding reactions at A and D, and (c) plot 8 versus the coefficient of friction. Hints: i. From Three Force Body Diagram and T -CD 120° T -A6 19 ROP/ B E C F 60° 200 N ii. TAB = ... TCD iii. Use law of Cosines and Sines to Solve for .Tcp then solve for in terms of coefficient of friction.