Question

n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an angle θ, of 60° with the floor. A man weighing 864 N climbs slowly up the ladder When he has climbed to a point that is 7.8 m from the base of the ladder, the ladder starts to slip. What is the coefficient of static friction between the floor and the ladder?

Answer 1

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

Answer 2

Answer:

friction coefficient will be given as

$\mu = 0.35$

Explanation:

By force balance in x direction we can write

$N_2 = \mu N_1$

force balance in y direction we can write

$N_1 = 400 + 864$

$N_1 = 1264 N$

now we will have

$\mu(1264) = N_2$

Now by torque balance about bottom most contact point

$N_2(12sin60) = 864(7.8cos60) + 400(6cos60)$

$N_2(10.4) = 3369.6 + 1200$

$N_2 = 439.4 N$

now from above equation we have

$\mu (1264) = 439.4$

$\mu = 0.35$