A house painter stands 3 m above the ground on a 4.9-m-long ladder that leans against the wall at a point 4.7 m above the ground. The painter weighs 697 N and the ladder weighs 180 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.
magnitude
N
direction
away from the wall towards the wall
You have two forces: Vertical force, which is the weight of the ladder and painter, which is the vertical force the driveway supplies.
The other force is horizontal, friction, which keeps the ladder from slipping. Sum moments about the upper end of the ladder, the two weights, the vertical force of the driveway, and the horizontal force on the driveway. You will get the horizontal force directly in the equation.
To find the force of friction that the driveway exerts on the bottom of the ladder, we need to consider the forces acting on the ladder. There are three forces involved: the weight of the painter, the weight of the ladder, and the normal force exerted by the driveway.
1. Weight of the painter: The weight of the painter can be calculated using the formula W = mg, where m is the mass and g is the acceleration due to gravity. We are given that the weight of the painter is 697 N.
2. Weight of the ladder: Similarly, the weight of the ladder can also be calculated using the same formula. We are given that the weight of the ladder is 180 N.
3. Normal force exerted by the driveway: The normal force is the force exerted by a surface perpendicular to it to support the weight of an object resting on it. In this case, the normal force is exerted by the driveway on the bottom of the ladder. Since the ladder is in equilibrium (not accelerating), the normal force will be equal in magnitude but opposite in direction to the sum of the other two forces.
To calculate the normal force, we can use the principle of moments. Moments or torques are forces that tend to cause rotation. The sum of the moments acting on the ladder must be zero in order for it to be in equilibrium.
Let's assume the direction away from the wall as the positive direction.
The moment due to the weight of the painter is (3 m - 4.7 m) * 697 N (since the center of gravity of the painter is 3 m above the bottom of the ladder).
The moment due to the weight of the ladder is (0 m - 4.7 m) * 180 N (since the center of gravity of the ladder is at the bottom).
If the force of friction is F and the distance from the bottom of the ladder to the point where the force of friction acts is 4.7 m, the moment due to the frictional force is (4.7 m) * F.
Since the ladder is in equilibrium, the sum of these moments must be zero:
(3 m - 4.7 m) * 697 N + (0 m - 4.7 m) * 180 N + (4.7 m) * F = 0
Simplifying the equation, we get:
(-1.7 m) * 697 N - (4.7 m) * 180 N + (4.7 m) * F = 0
Solving for F, we find:
F = ((1.7 m) * 697 N + (4.7 m) * 180 N) / (4.7 m)
F ≈ 124.3 N
Therefore, the force of friction that the driveway exerts on the bottom of the ladder is approximately 124.3 N, directed towards the wall.