EXAMPLE 8.7 Don't Climb the Ladder GOAL Apply the two conditions of equilibrium. (a) A ladder leaning against a frictionless wall. (b) A free-body diagram
EXAMPLE 8.7 Don't Climb the Ladder GOAL Apply the two conditions of equilibrium. (a) A ladder leaning against a frictionless wall. (b) A free-body diagram of the _, ladder. (c) Lever arms for the force of gravity and P. CD PROBLEM A uniform ladder 10.0 m long and weighing 50.0 N rests against a frictionless vertical wall as in Figure (a). If the ladder is just on the verge of slipping when it makes a 50.0 angle with the ground, find the coefficient of static friction between the ladder and ground. _, STRATEGY Figure (b) is the force diagram for the ladder. The first condition of equilibrium, 21F, = 0, gives two equations for three unknowns: the magnitudes of the static friction force f and the normal force n, both acting on the base of the ladder, and the magnitude of the force of the wall, P, acting on the top of the ladder. The second condition of equilibrium, 21', = 0, gives a third equation (for P), so all three quantities can be found. The definition of static friction then allows computation of the coefficient of static friction. SOLUTION Apply the first condition of equilibrium (1) ZFX = f P = 0 > f= P t the ladde\" (2) SF), = n - 50.0 N = 0 > n = 50.0 N Apply the second condition of 21', = If + In + T equilibrium, computing torques around the base of the ladder, with r grav + 1-P = D grav standing for the torque due to the ladder's 50.0-N weight. The torques due to friction and the 0 + 0 (50-0 N)(5-00 m) sin 40-0 4' P(10-0 m) sin 0 = normal force are zero about 0 because 50'0 their moment arms are zero. (Moment P = 21.0 N arms can be found from Figure (c).) From Equation (1), we now have 21.0 N = f= fS max = ,usn = ,us(50.0 N) f= P = 21.0 N. The ladder IS on the #5 = 21.0 N = 0-420 verge of slipping, so write an 50.0 N expression for the maximum force of static friction and solve for us. LEARN MORE REMARKS Note that torques were computed around an axis through the bottom of the ladder so that _, only P and the force of gravity contributed nonzero torques. This choice of axis reduces the complexity of the torque equation, often resulting in an equation with only one unknown. QUESTION If a 50.0 N monkey hangs from the middle rung, the coefficient of static friction would be doubled, halved, or unchanged? 0 doubled O halved O unchanged PRACTICE IT Use the worked example above to help you solve this problem. A uniform ladder I = 10.0 m long and weighing W = 50.4 N rests against a smooth vertical wall. If the ladder is just on the verge of slipping when it makes a 9 = 48.0 angle with the ground, find the coefficient of static friction between the ladder and ground. EXERCISE HINTS: GETTING-STARTED | I'M STUCK! If the coefficient of static friction is 0.398, and the same ladder makes a 64.0 angle with respect to the horizontal, how far along the length of the ladder can a 71.4-kg painter climb before the ladder begins to slip? |:lm
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