GOAL Apply the two conditions of equilibrium PROBLEM A uniform ladder 10.0 m lon
ID: 1464202 • Letter: G
Question
GOAL Apply the two conditions of equilibrium PROBLEM A uniform ladder 10.0 m long and weighing 50.0 N rests against a smooth 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 10 m 50 N 50 N 50° d. 0 (a) A ladder leaning against a frictionless wall. (b) A free- body diagram of the ladder. (c) Lever arms for the force of STRATEGY Figure (b) is the force gravity and P diagram for the ladder. The first condition of equilibrium, XF-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, = 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 to the ladder (2) 5-= n-50.0 N = 0 n = 50.0 N Apply the second condition of equilibrium, computing torques around the base of the ladder, with grav standing for the torque due to the ladder's 50.0-N weightExplanation / Answer
along horizantal
f - P = 0
P = f = u*N
along vertical
N - W = 0
N = W
torque about point
0 + 0 - (w*L/2*costheta) + (p*L*sintheta) = 0
(41.8*5*sin45) = (u*41.8*10*sin45)
u = 0.5
++++
along horizantal
f - P = 0
P = f = u*N
along vertical
N - W1 - w2 = 0
N = W1 + w2 = 41.8+69.5 = 111.3 N
torque about point o = 0
0 + 0 - (w1*L/2*cos61) - (w2*l*cos61) + (p*L*sin61) = 0
(41.8*5*sin45) +(59.5*l*cos61) = (0.364*111.3*10*sin61)
l = 7.16 m <<------answer
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.