(a) sum of torques = zero so mg(10 cos(theta) = Fc(20 sin(theta))
(mass is assumed at center)
sum of forces = zero so Fay = mg (really the normal component of Fa).
and Fc = Fax = (mu)Fay
So Fc = (mu)mg (if you assume the 18kg given in part b = .5*18*9.8)
Fa has x and y components, Fax = Fc =(mu)mg, Fay = mg. By pythagorean Fa = sqrt(mg^2 + (mu)mg^2)
(b) For no slipping the torques must be zero:
mg(10cos(theta))= (mu) mg (20 sin(theta))
(oddly the mass is not needed for part b)
tan(theta) = 10/(.5 * 20)
theta = 45o
A ladder ABC of length 20m is leaning against a wall at an angle from the horizontal. There is a roller at the top of the ladder that rests against the vertical wall. The coefficient of friction between the ladder and the horizontal surface at A is 0.5. Determine:
(a) the reactions at A and C;
(b) the smallest angle that the ladder alone can make with the wall without slipping if the ladder has a mass of 18kg.
1 answer
1 answer