Question

A uniform stationary ladder of length L = 2.7 m and mass M = 14 kg leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. The coefficient of static friction between floor and ladder is μ = 0.39. The ladder makes an angle θ = 55∘ with respect to the floor. A painter of mass 8M stands on the ladder a distance d from its base.(a) Find the magnitude of the normal force N, in newtons, exerted by the floor on the ladder.(b) Find an expression for the magnitude of the normal force NW exerted by the wall on the ladder.(c) What is the largest distance up the ladder (d) max, in meters, that the painter can stand without the ladder slipping?

Answer 1

Answer:

A) 1236 N

B) Nw = µ_s•N

C) d_max = 1.525 m

Explanation:

From the question, "smooth vertical wall" means that there is no friction there and thus the only vertical forces are the weights of the ladder and painter and the normal force at the floor.

a) Mass of ladder = 14 kg

Mass of painter = 8M = 8 * 14 = 112 kg

Thus, magnitude of normal force is;

N = total mass x acceleration due to gravity = (14 + 112)9.8

N = 1236 N

(b) Sum of the moments about the base of the ladder:

ΣH = 0

Nw - µ_s•N = 0

Nw = µ_s•N

c) Since they are the only two horizontal forces in play, we know that

Nw = Ff where Ff is the friction force at the floor.

Ff = µ_s*N = 0.39 × 1236

Ff = 482.04 N

So, to find maximum distance painter can stand without slipping, we'll use the formula ;

Nw(Lsinθ) = (Mgcosθ)(L/2) + (8Mgcosθ * d_max)

Plugging in the relevant values, we have;

482.04(2.7*sin55) = ((14 × 9.8cos55)*(2.7/2)) + (8*14*9.8cos55 * d_max)

1066.133 = 106.238 + 629.5575*d_max

629.5575*d_max = 1066.133 - 106.238

629.5575*d_max = 959.895

d_max = 959.895/629.5575

d_max = 1.525 m