Question

A house painter stands 3.00 m above the ground on a 5.00-m-long ladder that leans against the wall at a point 4.70 m above the ground. The painter weighs 676 N and the ladder weighs 107 N. Assuming no friction between the house and the upper end of the ladder, find the force of friction that the driveway exerts on the bottom of the ladder. If the force of friction is away from the wall, enter a negative value and if the force of friction is toward the wall, enter a positive value.

Answer 1

Answer:

The force of friction that the driveway exerts on the ladder ≈ -176.03 N

Please see attached force diagram created with MS Visio

Explanation:

The height of the house painter above the ground = 3.00 m

The length of the ladder = 5.00 m

The height above the ground at which the ladder leans on the wall = 4.70 m

The weight of the painter = 676 N

The weight of the ladder = 107 N

The angle of inclination of the ladder, arcsin(4.7/5) ≈ 70.05°

For equilibrium, the sum of moments at a point, ∑M = 0

Taking moment about point A gives;

4.7 × R = 107 × 2.5 × cos(arcsin(4.7/5)) + 676 × 3.0/(tan(arcsin(4.7/5)))

R = (107 × 2.5 × cos(arcsin(4.7/5)) + 676 × 3.0/(tan(arcsin(4.7/5))))/4.7 ≈ 176.03

The reaction force at the top of the ladder, R ≈ 176.03 N

For equilibrium, we have;

The sum of the horizontal forces, ∑Fₓ = 0

The reaction at the top of the ladder, R + The force of friction that the driveway exerts on the bottom of the ladder, $F_R$ = 0

R + $F_R$ = 0

∴ R = $-F_R$

R ≈ 176.03 N = $-F_R$

∴ $F_R$ = -176.03 N

Therefore, the force of friction that the driveway exerts on the bottom of the ladder, $F_R$ ≈ -176.03 N (acting away from the wall).