PROBLEM 6: A 6-m long ladder weighing 600 N is shown. It is required to determine the horizontal force P applied at C to hold the ladder from sliding. The coefficient of friction between the ladder and the contact surfaces at A and B is 0.20. (a) Determine the reaction at A. (b) Determine the reaction at B. (c) Determine the required force P. 1.5 m P A 30° 4.5 m
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- The uniform ladder of weight W is raised slowly by applying a vertical force P to the rope at A. Show that P is independent of the angle .The figure shows a three-pin arch. Determine the horizontal component of the pin reaction at A caused by the applied force P.A 1.1-kg disk A is placed on the inclined surface. The coefficient of static friction between the disk and the surface is 0.35. Is the disk in equilibrium if P=5.5N and =30?
- Find the magnitude of the pin reaction at B caused by the weight W=80lb. Neglect the weights of the members.The homogeneous 80-kg sign is suspended from a ball-and-socket joint at O, and cables AD and BC. Determine the forces in the cables.The block of weight W is pulled by the force P inclined at the angle to the horizontal. Find the smallest force P and the corresponding angle that would cause impending sliding of the block. The angle of static friction between the block and the ground is s.
- Determine the tension TB in Sample Problem 5.6 using one scalar equilibrium equation.point B. Problem 4.9 Figure 4.56 illustrates a person who is trying to pull a block on a horizontal surface using a rope. The rope makes an angle 0 with the horizontal. If W is the weight of the block and u is the coefficient of maximum Answer: T Determ force at pe friction between the bottom surface of the block and horizontal surface, show that the magnitude P of minimum force the person must apply in order to overcome the frictional and gravitational effects (to start moving the block) is P 3= cos e+ u sin 0 Fig. 4.58 Probl Fig. 4.56 Problem 4.9 Problem 4. 12 beam hinged to is / 4.5 m and Type here to searchQ.3) The coefficient of friction between the 100 lb block (shown in figure below) and the incline plane is 0.25 and that between the cord and cylindrical support is 0.3. Determine the range of cylinder weight W for which the system shown below will be in equilibrium. p = 0.3 100 lb p = 0.25 25° W
- Sample Problem 1. An engine with weight w hangs from a chain that is linked at a point O to two other chains, one is connected to the ceiling and the other one to the wall. Find the tension in each of the three chains, assuming that w is given and the weights of the chains are negligible. Fren ty agram of ergine h m of ring O Fre toty Solution: Ti is equal to weight w. The horizontal and slanted chains don't exert forces on the engine itself but they are connected on the ring where the three chains join. The weight of the ring itself is negligible. T, Ta, and Ty are the magnitudes of the forces while their directions are shown in the free-body diagram. Take note that the downward force with magnitude Ti of the chain acting on the engine are not an action-reaction pair. Using the x and y-components, we find that EFx 0, Ta cos 70 + (-Ta)-0 EF,-0, Ts sin 70' + (-T)-0 Because T-w, we can rewrite the second equation as: T3-T1/sin 70- w/ sin 70-0.940w We can now use this equation in the…2. A crate is kept in equilibrium by several frictionless rope-and-pulley arrangements shown in Figure 2. The rope can support a maximum tension force, T, of 100 N, and the mass of each pulley in the system is 1 kg. Draw the free-body diagrams for each arrangement and determine which one(s) can balance the heaviest crate and calculate the weight of that crate. TEND (a) (b) (c) (el) (e) Figure 2 A crate balanced by several rope-and-pulley arrangements.GOAL Apply the equilibrium conditions to the human body. Humerus PROBLEM A 50.0-N (11-lb) bowling ball is held in a person's hand with the forearm horizontal, as in Fig- ure 8.13a. The biceps muscle is attached 0.030 0 m from the joint, and the ball is 0.350 m from the joint. Find the upward force F exerted by the biceps on the forearm (the ulna) and the downward force R exerted by the humerus on the forearm, acting at the joint. Neglect the weight of the forearm and slight deviation from the vertical of the biceps. 50.0 N -Biceps Ulna 0.030 0 m 50.0 N 0.0300 m R 0.350 m -0.350 m-