Question

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A uniform ladder rests against a frictionless wall at an angle of -46.0° with respect to the ground. If the weight of the ladder is W-280 N, the length is L-7.5m, and the coefficient of static friction between the ladder and the ground is μs-0.75 determine the distance a man with a weight of WM-825 N can climb along the ladder before the ladder begins to slip. 8.77 since the ladder is in equilibrium up to the instant it starts to sto (which is also the instant the man has travelled as tar up the ladder posible before t beans to iipk we expect to use both the first and second condition of equilibrium, Since the second condition of equilibrium involves torque, can you write basic definition of torque in terms of the force causing the torque and the point of application of this force? Where is a good point about which to determine the torques? See if you can write a second condition of equilibrium statement that involives the distance we have been requested to determine. See if you can write a first condition of equilibrium statement for both vertical and horizontal forces that will allow you to eliminate any unknown quantities and then determine the distance the man can travel up the ladder before it begins to slip. m

Answer 1

By the given data The force is equal to the friction force f_t = 0.75 (280 + 825) = 0.75 times 1105 = 828.75 N F_t = 828.75 N Weight of ladder counter clockwise Torque = 140 times 1/2 times 8.5 times cos 46 = 140 times 8.5 times 0.6946 = 729.33 Weight of the Man Clockwise Torque = 829.75 times 7.5 times sin 46 degree = 828.75 times 7.5 times 0.71937 = 4470.89 w m \r\n clockwise torque 825 x d x cos 46 = 4470.89 - 729.33 d = 4470.89 - 729.33/825 times cos 46 degree = 3741.56/825 times 0.6946 3741.56/573.045 = 6.529 The ladder begins to slip = 6.529 m